datavorous/entrance-exam-dataset huggingface dataset:
A block of mass 100 kg slides over a distance of 10 m on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4 , then the work done against friction (in J) is:

Options:
A 4200
B 3900
C 4000
D 4500

The correct answer is:
4000

Solution:
Given m=100 kg

s=10 m

μ=0.4

The frictional force is given by

f=μmg=0.4×100×10=400 N

So, the work done by the force can be calculated as follows:

W=f×s=400×10=4000 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F=x N as shown in figure.

The value of x=_____.

Options:

The correct answer is:
12

Solution:
Balancing the forces in horizontal direction Nsinθ=F

And balancing the forces in vertical direction Ncosθ=mg

⇒tanθ=Fmg⇒F=mgtanθ

F=0.2×10×3=12 N. Therefore, x=12

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A block of mass 200 kg is being pulled by some men, up an inclined plane of inclination 45° and coefficient of friction 0.5, as shown in the figure. If each man can apply a maximum force of 500 N, then the minimum number of men required to just slide the block, up the plane is

Options:
A 10
B 7
C 5
D 9

The correct answer is:
5

Solution:
Total force required
F=mgsinâ¡Î¸+f
=mgsinâ¡Î¸+Î¼sR
=mgsinâ¡Î¸+Î¼s mgcosâ¡Î¸
=mg(sinâ¡Î¸+Î¼scosâ¡Î¸)
=200Ã9.8(sinâ¡45Â°+0.5cosâ¡45Â°)
F=200Ã9.8Ã322
The number of men required will be
200Ã9.8Ã3(500)Ã22=4.1

Topic: Physics, Laws of Motion, JEE Main
A block of mass m=0.1 kg is released from a height of 4 m on a curved smooth surface. On the horizontal surface, path AB is smooth and path BC is rough with a coefficient of friction Î¼=0.1. If the impact of the block with the vertical wall at C is perfectly elastic, the total distance covered by the block on the horizontal surface before coming to rest will be (take g=10 m s-2)

Options:
A 29 m
B 49 m
C 59 m
D 109 m

The correct answer is:
59 m

Solution:
Total KE dissipated against friction =mgh=4 J
Work done against friction in one trip across BC=mgμ×2=0.2 J
∴ Total number of trips across BC=40.2=20
∴ Total distance =20×BC+19×AB=59 m

Topic: Physics, Work Power Energy, JEE Main
A block of mass M1 is constrained to move along with a moveable pulley of mass M2 which is connected to a spring of force constant k, as shown in the figure. If the mass of the fixed pulley is negligible and friction is absent everywhere, then the period of small oscillations of the system is

Options:
A 2π4M1+M2k
B 2π4M2+M1k
C 4π2M1+M2k
D π4M1-M2k

The correct answer is:
2π4M1+M2k

Solution:
If the mass M1 is displaced by x and has speed v, spring will extend by x/2 and M2 will have speed v/2. Energy of this system can be written as
-M1gx+12M1v2+M2gx2+12M2vâ¡22+12kx22=const.
-M1gx+12M1v2+M22gx+18M2v2+18kx2=const.
Differentiate wrt time
-M1gv+M1vdvdt+M2g2v+14M2vdvdt+18kx.v=0
â-M1g+M1dvdt+M2g2+14M2dvdt+18kx=0
M1+M24dvdt=-k4x+M1g-M2g2
a=dvdt=-k4M1+M2x+22M1-M2g4M1+M2
Here,Ï2=k4M1+M2âT=2ÏÏ=2Ï4M1+M2k

Topic: Physics, Oscillations, JEE Main
A block of mass m=0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance x2 from the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity 3 m s-1. The total initial energy of the spring is:

Options:
A 0.6 J
B 0.8 J
C 1.5 J
D 0.3 J

The correct answer is:
0.6 J

Solution:
By mechanical energy conservation between compression positions x and x2
12kx2=12kx22+12mv2
12kx2-12kx24=12mv2
12kx234=12mv2
v=3kx24m=3kmx2
On collision with a block at rest
∵ Velocities are exchanged ⇒ elastic collision between identical masses.
∴v=3=3kmx2
⇒6=3km x
⇒x=6m3k
∴ The initial energy of the spring is
U=12k x2=12k×36m3k=6m
U=6×0.1=0.6 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2015 (10 Apr Online)
A block of mass m=0.1 kg is connected to a spring of unknown spring constant k. It is compressed to a distance x from its equilibrium position and released from rest. After approaching half the distance x2 from the equilibrium position, it hits another block and comes to rest momentarily, while the other block moves with velocity 3 m s-1. The total initial energy of the spring is:

Options:
A 0.6 J
B 0.8 J
C 1.5 J
D 0.3 J

The correct answer is:
0.6 J

Solution:
By mechanical energy conservation between compression positions x and x2
12kx2=12kx22+12mv2
12kx2-12kx24=12mv2
12kx234=12mv2
v=3kx24m=3kmx2
On collision with a block at rest
∵ Velocities are exchanged ⇒ elastic collision between identical masses.
∴v=3=3kmx2
⇒6=3km x
⇒x=6m3k
∴ The initial energy of the spring is
U=12k x2=12k×36m3k=6m
U=6×0.1=0.6 J

Topic: Physics, Work Power Energy, JEE Main
A block of mass m=1 kg moving on a horizontal surface with speed u=2 m s-1 enters a rough horizontal patch ranging from x = 0.10 m to x = 2.00 m. If the retarding force fr on the block in this range is inversely proportional to x over this range i.e.
fr=-kx for 0.10<x<2.00
fr=0 for x<0.10â and x>2.00
If k=0.5 J then, the speed of this block ( in m s-1 ) as it crosses the patch is ( use lnâââ20=3 )

Options:

The correct answer is:
1

Solution:
W=â«frdx=-klogeâ¡20.1=-1.5 J
â´W=âK
12Ã1Ãv2-12Ã1Ã4=-1.5

âv=1 m s-1

Topic: Physics, Work Power Energy, JEE Main
A block of mass m=1 kg slides with velocity v=6 m s-1 on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about O and swings as a result of the collision making angle θ before momentarily coming to rest. if the rod has mass M=2 kg, and length ℓ=1 m, the value of θ is approximately(take g=10 m s-2

Options:
A 63°
B 55°
C 69°
D 49°

The correct answer is:
63°

Solution:
Angular momentum

mvl=ml2+2ml23ω

mvl=53ml2ω

ω=3v5l

12Iω2=2mgl21-cos θ+mgl1-cos θ

1253ml29v225l2=2mgl 1-cos θ

35×2mv2=2mgl1-cosθ

310×362×10=1-cos θ

1-2750=cos θ

cos θ=2350=0.46

θ=cos-10.46=63°

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A block of mass M=2 kg with a semicircular track of radius R=1.1 m rests on a horizontal frictionless surface. A uniform cylinder of radius r=10 cm and mass m=1.0 kg is released from rest from the top point A. The cylinder slips on the semicircular frictionless track. The speed of the block when the cylinder reaches the bottom of the track at B is g=10 m s-2

Options:
A 103 m s-1
B 43 m s-1
C 52 m s-1
D 10 m s-1

The correct answer is:
103 m s-1

Solution:
Let the speed of the block be v. Then from conservation of momentum, the velocity of the cylinder will be 2v in the opposite direction m=M2
Now from conservation of energy
mgh=12Mv2+12m2v2, h=R-r=1 m
so v=103 m s-1

Topic: Physics, Work Power Energy, JEE Main
A block of mass m=2 kg is placed on a plank of mass M=10 kg, which is placed on a smooth horizontal plane, as shown in the figure. The coefficient of friction between the block and the plank is Î¼=13 . If a horizontal force F is applied on the plank, then the maximum value of F for which the block and the plank move together is g=10 m/s2

Options:
A 50 N
B 40 N
C 120 N
D 100 N

The correct answer is:
40 N

Solution:
a=FM+m assuming no slipping between m and M

This means friction between the block and plank is

f=ma=mFm+M

f=mFm+M≤μmg

⇒F≤M+mg3

⇒F≤40 N

Topic: Physics, Laws of Motion, JEE Main
A block of mass m=10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed v, that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely falling object were to acquire speed v10 after being dropped from height H, then neglecting energy losses and taking g=10 m s-2, the value of H is close to

Options:
A 0.2 km.
B 0.5 km.
C 0.4 km.
D None of these.

The correct answer is:
None of these.

Solution:
By the conservation of linear momentum,
m1v=m+m1v1
⇒v1=m1vm+m1 ...(1)
After collision by work energy theorem, we have
Wfriction=∆k
⇒-μm+m1gx=-12 m+m1v12
⇒μgx=12v12
⇒0.05×10×2=12m1 vm+m12
⇒5100×10×4=5010002×v210+ 5010002
⇒2=1400v2×4002012

v2=2012 2
v=2×201
For a freely falling body,
v'=2gH

v10=2gH

2×20110=2×gH
2012100=10H

H=20121000=40 m
=0.04 km

Topic: Physics, Work Power Energy, JEE Main
A block of mass m=10 kg rests on a horizontal table. The coefficient of friction between the block and the table is 0.05. When hit by a bullet of mass 50 g moving with speed v, that gets embedded in it, the block moves and comes to stop after moving a distance of 2 m on the table. If a freely falling object were to acquire speed v10 after being dropped from height H, then neglecting energy losses and taking g=10 m s-2, the value of H is close to

Options:
A 0.2 km.
B 0.5 km.
C 0.4 km.
D None of these.

The correct answer is:
None of these.

Solution:
By the conservation of linear momentum,
m1v=m+m1v1
⇒v1=m1vm+m1 ...(1)
After collision by work energy theorem, we have
Wfriction=∆k
⇒-μm+m1gx=-12 m+m1v12
⇒μgx=12v12
⇒0.05×10×2=12m1 vm+m12
⇒5100×10×4=5010002×v210+ 5010002
⇒2=1400v2×4002012

v2=2012 2
v=2×201
For a freely falling body,
v'=2gH

v10=2gH

2×20110=2×gH
2012100=10H

H=20121000=40 m
=0.04 km

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2015 (10 Apr Online)
A block of mass m and charge q is released on a long smooth inclined plane. Magnetic field B is constant, uniform and out of the plane of the paper as shown. Find the time from the start when blocking lose contact with the surface.

Options:
A mâcosÎ¸qB
B mâcosec Î¸qB
C mâcotÎ¸qB
D mâtanÎ¸qB

The correct answer is:
m cotθqB

Solution:
Block will lose contact with surface when force due to magnetic field will become equal to mgcosθ
qvB=mgcosθ
v=mgcosθqB=u+at ( along inclined plane )
v=mgcosθqB=(g sinθ)t
t= m cotθqB

Topic: Physics, Magnetic Effects of Current, JEE Main
A block of mass m attached to a massless spring is performing oscillatory motion of amplitude 'A' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become ƒA. The value of ƒ is:

Options:
A 12
B 1
C 12
D 2

The correct answer is:
2

Solution:
Using mometum conservation

m1v1=m2v2mAω=m2A2ω2mAkm=m2A2km/2A2=2A⇒f=2

Topic: Physics, Oscillations, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A block of mass $m$ initially at rest is dropped from a height $h$ on to a spring of force constant $k$. the maximum compression in the spring is $x$ then

Options:
A $m g h=\frac{1}{2} k x^2$
B $m g(h+x)=\frac{1}{2} k x^2$
C $m g h=\frac{1}{2} k(x+h)^2$
D $m g(h+x)=\frac{1}{2} k(x+h)^2$

The correct answer is:
$m g(h+x)=\frac{1}{2} k x^2$

Solution:
From Energy conservation, It is known that
Change in gravitational potential energy \(=\) Elastic potential energy stored in compressed spring
\(\begin{aligned}
& \mathrm{E}_{\mathrm{g}}=\mathrm{mg}(\mathrm{h}+\mathrm{x}) \\
& \mathrm{E}_{\mathrm{s}}=\frac{1}{2} \mathrm{kx}^2 \\
& \Rightarrow \mathrm{mg}(\mathrm{h}+\mathrm{x})=\frac{1}{2} \mathrm{kx}^2
\end{aligned}\)

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A block of mass m is attached to a cart of mass 4m through spring of spring constant k as shown in the figure. Friction is absent everywhere. The time period of oscillations of the system, when spring is compressed and then released, is

Options:
A πmk
B 2πmk
C 2π5m2k
D 4πm5k

The correct answer is:
4πm5k

Solution:
It is equivalent to a system as shown below,
T=2πmreducedk
∴ T=2πm×4m5mk
= 2π4m5k = 4πm5k

Topic: Physics, Oscillations, JEE Main
A block of mass m is attached to one end of a light string which is wrapped on a disc of mass 2m and radius R. The total length of the slack portion of the string is l. The block is released from rest. The angular velocity of the disc just after the string becomes taut is

Options:
A glR2
B gl2R2
C 2gl3R2
D 3gl2R2

The correct answer is:
gl2R2

Solution:
Using the principle of conservation of angular momentum,

m2gl×R=2m R22+mR2×ω

⇒ω=gl2R2

Topic: Physics, Rotational Motion, JEE Main
A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $\mathrm{k}$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be :

Options:
A $\mathrm{Mg} / 2 \mathrm{k}$
B $\mathrm{Mg} / \mathrm{k}$
C $2 \mathrm{Mg} / \mathrm{k}$
D $4 \mathrm{Mg} / \mathrm{k}$

The correct answer is:
$2 \mathrm{Mg} / \mathrm{k}$

Solution:
Loss in grav. $\mathrm{PE}=$ gain in spring $\mathrm{PE}$
At maximum elongation
$$
\begin{aligned}
\mathrm{Mgx} & =\frac{1}{2} \mathrm{kx}^2 \\
\mathrm{x} & =\frac{2 \mathrm{Mg}}{\mathrm{k}}
\end{aligned}
$$

Topic: Physics, Work Power Energy, JEE Main
A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $\mathrm{k}$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be :

Options:
A $\mathrm{Mg} / 2 \mathrm{k}$
B $\mathrm{Mg} / \mathrm{k}$
C $2 \mathrm{Mg} / \mathrm{k}$
D $4 \mathrm{Mg} / \mathrm{k}$

The correct answer is:
$2 \mathrm{Mg} / \mathrm{k}$

Solution:
Loss in grav. $\mathrm{PE}=$ gain in spring $\mathrm{PE}$
At maximum elongation
$$
\begin{aligned}
\mathrm{Mgx} & =\frac{1}{2} \mathrm{kx}^2 \\
\mathrm{x} & =\frac{2 \mathrm{Mg}}{\mathrm{k}}
\end{aligned}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2009 (Mains)
A block of mass \( M \) is attached to three springs, \( A, B \& C \) having force constants \( K, K \) and \( 3 K \) as shown in the figure. If the block is slightly pushed against spring \( C \). Then find the frequency of oscillations,

Options:
A \( f=2 \pi \sqrt{\frac{M}{K}} \)
B \( f=\frac{1}{2 \pi} \sqrt{\frac{K}{M}} \)
C \( f=\frac{1}{2 \pi} \sqrt{\frac{2 K}{M}} \)
D \( f=\frac{1}{2 \pi} \sqrt{\frac{3 K}{M}} \)

The correct answer is:
\( f=\frac{1}{2 \pi} \sqrt{\frac{2 K}{M}} \)

Solution:
Suppose the common point is taken x distance along point C.

From the diagram, we can see that extension in the spring A and B ,

x1=xcos45°=x2.

Extension in spring C,
x2=x

Force developed in spring A and B,

F1=kx1=kx2.

Force developed in every spring C,
F2=kx

Sum of components of all these forces in the direction of OC will provide restoring force,

Fnet==-2×kx2cos45°+kx=2×kx2+kx

Fnet=-2kx.

Using newton's second law-

Fnet=ma=-2kx

ma=-2kx

a=-2kmx

For SHM,
a=-ω2x=-2πT2x

∴ 2πT2=2km

⇒T=2πm2k

Topic: Physics, Oscillations, JEE Main
A block of mass m is connected rigidly with a smooth wedge (plank) by a light spring of stiffness k. If the wedge is moved with constant velocity v0, then find the work done by the external agent until the maximum compression of the spring.

Options:
A m⁡v⁡ 0 2
B 2 m&ApplyFunction; v&ApplyFunction; 0 2
C 3 m⁡ v⁡ 0 2
D 2 m⁡ v⁡ 0 3

The correct answer is:
m⁡v⁡ 0 2

Solution:
Let us take wedge + spring + block as a system. The forces responsible for performing work are spring force kx (â) and the external force F(â).
Work Energy theorem for block + spring + plank relative to ground :
Applying work-energy theorem, we have Wext + Wsp = ÎK
where Wsp the total work done by the spring on wedge and block - 1 2 k x 2 and ÎK = change in KE of the block (because the plank does not change its kinetic energy)
Then, W ext = 1 2 k x 2 + Î K
As the block was initially stationary and it will acquire a velocity v0 equal to that of the plank at the time of maximum compression of the spring, the change in kinetic energy of the block relative to ground is
Î K = 1 2 mâ¡ vâ¡ 0 2
Substituting ÎK in the above equation, we have
W ext = 1 2 K x 2 + 1 2 mâ¡ vâ¡ 0 2 ...(i)
Work Energy theorem for block + spring + plank relative to the plank Wext + Wsp = ÎK The plank moves with constant velocity, there is no pseudo-force acting on the block. Wext = 0 Then the net work done on the system (block + plank), due to the spring, can be given as
W SP = - 1 2 k x 2
As the relative velocity between the observer (plank) and block decreases from v0 to zero at the time of maximum compression of the spring, the change in kinetic energy of the block is Î K = - 1 2 mâ¡ vâ¡ 0 2 .
Substituting Wsp and â³K in above equation
-1 2 kâ¡ x 2 = -1 2 mâ¡ vâ¡ 0 2 (ii)
From (i) & (ii)
W ext =12mv02+12mv02=mv02

Topic: Physics, Work Power Energy, JEE Main
A block of mass ' $\mathrm{m}$ ' is connected to another block of mass ' $\mathrm{M}$ ' by a spring (massless) of spring constant ' $\mathrm{k}$ '. The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest and the spring is unstretched. Then a constant force '$\mathrm{F}$' starts acting on the block of mass '$\mathrm{M}$' to pull it. Find the force on the block of mass '$\mathrm{m}$'

Options:
A
$\frac{\mathrm{mF}}{\mathrm{M}}$

B
$\frac{\mathrm{(M+m) F}}{\mathrm{m}}$

C
$\frac{\mathrm{m F}}{\mathrm{(m+M)}}$

D
$\frac{\mathrm{MF}}{(\mathrm{m}+\mathrm{M})}$

The correct answer is:
$\frac{\mathrm{m F}}{\mathrm{(m+M)}}$

Solution:
$\mathrm{Kx}=\mathrm{ma}=\frac{\mathrm{mF}}{\mathrm{m}+\mathrm{M}}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2007
A block of mass m is gently placed over a massive plank moving horizontally over a smooth surface with velocity 10 m s-1. The coefficient of friction between the block and the plank is 0.2. The distance travelled by the block till it slides on the plank is g=10 m s-2

Options:
A 10Â m
B 15Â m
C 25 m
D 35Â m

The correct answer is:
25 m

Solution:
Initially the block is not moving with respect to ground while plank is moving, it means, relative motion is there between the plank and block due to which friction (kinetic) starts acting. Initial velocity of block is zero and final velocity is 10 m/s and let s is the distance travelled by the block over the ground, then:-
m1022-m022=0.2×m×10×s
⇒50=2s⇒s=25 m

Topic: Physics, Work Power Energy, JEE Main
A block of mass $\mathrm{m}$ is in contact with the cart $\mathrm{C}$ as shown in the figure.
The coefficient of static friction between the block and the cart is $\mu$. The acceleration $\alpha$ of the cart that will prevent the block from falling satisfies

Options:
A $\alpha>\frac{m g}{\mu}$
B $\alpha>\frac{g}{\mu \mathrm{m}}$
C $\alpha \geq \frac{g}{\mu}$
D $\alpha < \frac{g}{\mu}$

The correct answer is:
$\alpha \geq \frac{g}{\mu}$

Solution:
When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left. This force $(m \alpha)$ is action force by a block on cart Now, block will remain
static w.r.t. cart, if frictional force $\mu \mathrm{R} \geq \mathrm{mg}$
$$
\begin{array}{crrr}
\Rightarrow & \mu \mathrm{m} \alpha & \geq \mathrm{mg} & \text { [as } \mathrm{R}=\mathrm{m} \alpha \text { ] } \\
\Rightarrow & \alpha & \geq \frac{g}{\mu} &
\end{array}
$$

Topic: Physics, Laws of Motion, JEE Main
A block of mass $\mathrm{m}$ is in contact with the cart $\mathrm{C}$ as shown in the figure.
The coefficient of static friction between the block and the cart is $\mu$. The acceleration $\alpha$ of the cart that will prevent the block from falling satisfies

Options:
A $\alpha>\frac{m g}{\mu}$
B $\alpha>\frac{g}{\mu \mathrm{m}}$
C $\alpha \geq \frac{g}{\mu}$
D $\alpha < \frac{g}{\mu}$

The correct answer is:
$\alpha \geq \frac{g}{\mu}$

Solution:
When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left. This force $(m \alpha)$ is action force by a block on cart Now, block will remain
static w.r.t. cart, if frictional force $\mu \mathrm{R} \geq \mathrm{mg}$
$$
\begin{array}{crrr}
\Rightarrow & \mu \mathrm{m} \alpha & \geq \mathrm{mg} & \text { [as } \mathrm{R}=\mathrm{m} \alpha \text { ] } \\
\Rightarrow & \alpha & \geq \frac{g}{\mu} &
\end{array}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2010 (Screening)
A block of mass m is kept on a platform which starts from rest with a constant acceleration g/2 upwards, as shown in the figure. Work done by normal reaction on block in time t is

Options:
A -mg2t28
B mÂ g2t28
C 0
D 3 m g2t28

The correct answer is:
3 m g2t28

Solution:
N-mg=mg2
N=3mg2
s=ut+12at2
=0+12g2t2
=g4 t 2
WN=3mg2×g4 t2
=3mg2t28=3mg2t28

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A block of mass m is kept on a rough horizontal surface and is attached with a massless spring of force constant k. The minimum constant force applied on the other end of the spring to lift the block is

Options:
A Mg
B Mg2
C 2Mg
D Mg(1 + k)

The correct answer is:
Mg2

Solution:
As W ext = Δ E = 1 2 Kx 2 ; F.x = 1 2 Kx 2 or F = K x 2
For lifting Kx = Mg or F = Mg 2

Topic: Physics, Work Power Energy, JEE Main
A block of mass m is kept on an inclined plane of a lift moving down with an acceleration of 2 m s-2. What should be the coefficient of friction for the block to move down with constant velocity relative to lift? The angle of inclination of the wedge is 30Â°. g=10 m s-2

Options:
A Î¼ = 1 3
B μ = 0 . 4
C μ = 0 . 8
D μ = 0 . 5

The correct answer is:
μ = 1 3

Solution:
geff=g-2=8 m s-2
m 8 sin 30 o =μ.m 8 cos 30 o ⇒μ=tan 30 o
⇒ m g - a sin θ = μ m g - a cos θ
⇒ μ = tan θ = tan 3 0 o = 1 3

Topic: Physics, Laws of Motion, JEE Main
A block of mass $m$ is lying on a rough inclined plane having an inclination $\alpha=\tan ^{-1}\left(\frac{1}{5}\right)$. The inclined plane is moving horizontally with a constant acceleration of $a=2 \mathrm{~ms}^{-2}$ as shown in the figure. The minimum value of coefficient of friction, so that the block remains stationary with respect to the inclined plane is (Take, $g=10 \mathrm{~ms}^{-2}$ )

Options:
A $\frac{2}{9}$
B $\frac{5}{12}$
C $\frac{1}{5}$
D $\frac{2}{5}$

The correct answer is:
$\frac{5}{12}$

Solution:
The block-plane system is shown in the figure,
So, from the above free body diagram (FBD), downward acceleration,
Divided the Eq. (iii) by $\sin \alpha$ and putting the values,
$$
\begin{aligned}
& \mu=\frac{10+2 \frac{1}{\tan \alpha}}{10 \frac{1}{\tan \alpha}-2}=\frac{10+2 \times 5}{10 \times 5-2} \\
& \mu=\frac{20}{48}=\frac{5}{12}
\end{aligned}
$$
Hence, the correct option is (b).

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 1)
A block of mass 'M' is moving on rough horizontal surface with momentum 'P'. The coefficient of friction between the block and surface is ' $\mu$ '. The distance covered by block before it stops is $[\mathrm{g}=$ acceleration due to gravity $]$

Options:
A $\frac{2 \mu \mathrm{Mg}}{\mathrm{P}}$
B $\frac{\mathrm{P}}{2 \mu \mathrm{Mg}}$
C $\frac{\mathrm{P}^{2}}{2 \mu \mathrm{M}^{2} \mathrm{~g}}$
D $\frac{2 \mu \mathrm{M}^{2} \mathrm{~g}}{\mathrm{P}^{2}}$

The correct answer is:
$\frac{\mathrm{P}^{2}}{2 \mu \mathrm{M}^{2} \mathrm{~g}}$

Solution:
Using kinematic relation,
$v^{2}=u^{2}-2 a s$
The momentum pis given by
$\begin{array}{l}
p=M u \\
u=\frac{P}{M}
\end{array}$
and acceleration is given as
$a=\mu g$
Substituting values in Eq. (i), we get
$\begin{array}{l}
0=\mathrm{u} 2-2 \mathrm{as}(\because f \in \text { alvelocity }, v=0) \\
0=\left(\frac{\mathrm{p}}{\mathrm{M}}\right)^{2}-2 \mu \mathrm{gs} \\
\left(\frac{\mathrm{p}}{\mathrm{M}}\right)^{2}=2 \mu \mathrm{gs} \\
\Rightarrow \mathrm{s}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}^{2} \mu \mathrm{g}}
\end{array}$

Topic: Physics, Laws of Motion, JEE Main
A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is Î¼, the power delivered by the external agent after a time t, from the beginning, is equal to

Options:
A ma2t
B μmgat
C Î¼ m(a+Î¼g) gt
D m(a+Î¼g)at

The correct answer is:
m(a+μg)at

Solution:
The total power divided by the external agent is mainly consumed in two parts:
(1) To move the body in the forward direction.
(2) To compensate the effect of the friction.
(i) Work done in moving the body
W1 = F . S
W1 = ma S …(1)
(ii) Work done agent friction
W2 = fk × S = μk R S = μk W S
W2 = μk mg S …(2)
Net work done W = W1 + W2
W = ma S + μk mg S
W = m (a + μ g) S
Now, according to 2nd equation of motion
S = u + 12 at2
Now the power divided:–
P=dwdt=ddtma+μg12at2
P = m(a + μ g) at

Topic: Physics, Laws of Motion, NEET