Laws of Motion — Page 4 | JEE Physics

Q31

A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms

-

1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is : [use g = 9.8 ms

-

2]

A 4.9 m

B 9.8 m

C 12.5 m

D 19.6 m

Correct Answer

Option B

Solution

S = {{{u^2}} \over {2a}} = {{{u^2}} \over {2(\mu g)}}

= {{{{(9.8)}^2}} \over {2 \times 0.5 \times (9.8)}}

= {{9.8} \over 1}

= 9.8

m

Q32

A monkey of mass

50 \mathrm{~kg}

climbs on a rope which can withstand the tension (T) of

350 \mathrm{~N}

. If monkey initially climbs down with an acceleration of

4 \mathrm{~m} / \mathrm{s}^{2}

and then climbs up with an acceleration of

5 \mathrm{~m} / \mathrm{s}^{2}

. Choose the correct option

\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)

.

A

T=700 \mathrm{~N}

while climbing upward

B

T=350 \mathrm{~N}

while going downward

C Rope will break while climbing upward

D Rope will break while going downward

Correct Answer

Option C

Solution

Tdown = 50 $\times$ (10 $-$ 4) = 50 $\times$ 6 = 300 N Tup = 50 $\times$ (10 + 5) = 50 $\times$ 15 = 750 N $\Rightarrow$ Rope will break while climbing up.

Q33

A bag is gently dropped on a conveyor belt moving at a speed of

2 \mathrm{~m} / \mathrm{s}

. The coefficient of friction between the conveyor belt and bag is

0.4

. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is : [Take

\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{-2}

]

A 2 m

B 0.5 m

C 3.2 m

D 0.8 ms

Correct Answer

Option B

Solution

Speed of conveyor belt, $v=2 \mathrm{~m} / \mathrm{s}$ Coefficient of friction between the conveyor belt and bag, $\mu=0.4$ Acceleration of bag due to slipping motion,

a=\mu g=0.4 \times 10=4 \mathrm{~m} / \mathrm{s}^2 \quad\left(\because g=10 \mathrm{~m} / \mathrm{s}^2\right)

If $s$ be the distance travelled by the bag on the belt during slipping motion, then $\ v^2 \ =u^2-2 a s$ $\Rightarrow \ 0=v^2-2 a s$ $\Rightarrow s \ =\dfrac{v^2}{2 a}=\dfrac{2^2}{2 \times 4}=0.5 \mathrm{~m}$

Q34

A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is

\theta

. The magnitude of the contact force will be :

A Mg

B

\mathrm{Mg} \cos \theta

C

\sqrt{\mathrm{Mg} \sin \theta+\mathrm{Mg} \cos \theta}

D

\operatorname{Mg} \sin \theta \sqrt{1+\mu}

Correct Answer

Option A

Solution

As the body is moving with constant velocity so forces acting on the body must be balanced.

$\Rightarrow$ Contact force from incline should balance weight of the body.

$\Rightarrow$ | Fcontact | = Mg

Q35

A block 'A' takes 2 s to slide down a frictionless incline of 30

^\circ

and length 'l', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to 45

^\circ

, the time taken by the block, to slide down the incline, will be approximately :

A 2.66 s

B 0.83 s

C 1.68 s

D 0.70 s

Correct Answer

Option C

Solution

{\theta _1} = 30^\circ ,\,{\theta _2} = 45^\circ

{a_1} = g\sin {\theta _1} = 5

m/s2,

{a_2} = g\sin {\theta _2} = 5\sqrt 2

m/s2

{{{t_1}} \over {{t_2}}} = {{\sqrt {{{2l} \over {{a_1}}}} } \over {\sqrt {{{2l} \over {{a_2}}}} }} = \sqrt {{{{a_2}} \over {{a_1}}}}

{{{t_1}} \over {{t_2}}} = {(2)^{1/4}}

{t_2} = {(2)^{3/4}}

\approx 1.68

s

Q36

The time taken by an object to slide down 45

^\circ

rough inclined plane is n times as it takes to slide down a perfectly smooth 45

^\circ

incline plane. The coefficient of kinetic friction between the object and the incline plane is :

A

1 - {1 \over {{n^2}}}

B

1 + {1 \over {{n^2}}}

C

\sqrt {1 - {1 \over {{n^2}}}}

D

\sqrt {{1 \over {1 - {n^2}}}}

Correct Answer

Option A

Solution

Smooth case:

a = g\sin 45^\circ = {9 \over {\sqrt 2 }}

{t_1} = \sqrt {{{2L} \over a}} = \sqrt {{{2L} \over {g/\sqrt 2 }}} = \sqrt {{{2\sqrt 2 L} \over g}}

..... (1) Rough case:

a = g\sin 45^\circ - \mu g\cos 45^\circ

= {g \over {\sqrt 2 }}(1 - \mu )

{t_2} = \sqrt {{{2L} \over a}} = \sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}}

..... (2) From (1) to (2) and

{t_1} = {{{t_2}} \over n}

we have

\sqrt {{{2\sqrt 2 L} \over g}} = {1 \over n}\sqrt {{{2\sqrt 2 L} \over {g(1 - \mu )}}} \Rightarrow \mu = 1 - {1 \over {{n^2}}}

Q37

A block of mass

m

slides down the plane inclined at angle

30^{\circ}

with an acceleration

\dfrac{g}{4}

. The value of coefficient of kinetic friction will be:

A

\dfrac{2 \sqrt{3}-1}{2}

B

\dfrac{\sqrt{3}}{2}

C

\dfrac{1}{2 \sqrt{3}}

D

\dfrac{2 \sqrt{3}+1}{2}

Correct Answer

Option C

Solution

$\mathrm{Mg} \sin 30^{\circ}-\mu \mathrm{mg} \cos 30^{\circ}=\mathrm{ma}$

\frac{g}{2}-\frac{\sqrt{3}}{2} \cdot \mu g=\frac{g}{4}

\begin{aligned} & \frac{\sqrt{3}}{2} \mu=\frac{1}{4} \\\\ & \mu=\frac{1}{2 \sqrt{3}} \end{aligned}

Q38

A body of mass

500 \mathrm{~g}

moves along

\mathrm{x}

-axis such that it's velocity varies with displacement

\mathrm{x}

according to the relation

v=10 \sqrt{x} \mathrm{~m} / \mathrm{s}

the force acting on the body is:-

A 166 N

B 5 N

C 25 N

D 125 N

Correct Answer

Option C

Solution

Given that the velocity of the body varies with displacement x according to the relation:

v = 10\sqrt{x}\,\mathrm{ms}^{-1}

To find the force acting on the body, we first need to find its acceleration, which can be obtained by differentiating the velocity with respect to time.

However, we don't have the velocity expressed as a function of time, but rather as a function of displacement.

To work around this, we will use the chain rule:

\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}

Now, differentiate the velocity with respect to displacement:

\frac{dv}{dx} = \frac{1}{2} \cdot 10 \cdot x^{-1/2} = 5x^{-1/2}

Recall that

\frac{dx}{dt}

is the velocity, so we have:

\frac{dv}{dt} = 5x^{-1/2} \cdot 10\sqrt{x} = 50

Thus, the acceleration is constant and equal to 50 m/s².

Now we can find the force acting on the body using Newton's second law:

F = ma

First, convert the mass from grams to kilograms:

m = \frac{500\,\mathrm{g}}{1000} = 0.5\,\mathrm{kg}

Now, calculate the force:

F = (0.5\,\mathrm{kg})(50\,\mathrm{ms}^{-2}) = 25\,\mathrm{N}

The force acting on the body is 25 N.

Q39

A cricket player catches a ball of mass

120 \mathrm{~g}

moving with

25 \mathrm{~m} / \mathrm{s}

speed. If the catching process is completed in

0.1 \mathrm{~s}

then the magnitude of force exerted by the ball on the hand of player will be (in SI unit) :

A 30

B 24

C 12

D 25

Correct Answer

Option A

Solution

The first step in solving this problem is to calculate the change in momentum of the ball when it is caught.

The change in momentum, or impulse, is the product of the mass of the ball and the change in velocity (as momentum is mass times velocity).

The ball is initially moving with a velocity of $v_i = 25 \mathrm{~m/s}$ before the catch and finally comes to rest with a velocity of $v_f = 0 \mathrm{~m/s}$ after the catch.

Since the ball is caught, the final velocity is zero.

The change in velocity

\Delta v = v_f - v_i = 0 - 25 = -25 \mathrm{~m/s}.

Remember that the direction of the force exerted by the ball on the hand will be opposite to the direction of the ball's initial motion.

The mass of the ball $m$ is given as $120 \mathrm{~g}$ which needs to be converted into kilograms to maintain SI units:

m = 120 \mathrm{~g} = 120 \times 10^{-3} \mathrm{~kg} = 0.12 \mathrm{~kg}.

Now we can calculate the change in momentum (impulse):

\Delta p = m \Delta v = 0.12 \mathrm{~kg} \times (-25 \mathrm{~m/s}).

Substituting the values we get:

\Delta p = 0.12 \times -25 = -3 \mathrm{~kg \cdot m/s}.

The negative sign indicates that the change in momentum is in the opposite direction of the ball's initial motion, which makes sense because the ball's velocity is reduced to zero.

The magnitude of the impulse is independent of the sign and is $3 \mathrm{~kg \cdot m/s}$ .

Impulse is also equal to the average force exerted on the ball times the time interval during which the force is exerted.

We can use the formula:

\Delta p = F_{avg} \Delta t

Where $F_{avg}$ is the average force and $\Delta t$ is the time interval of $0.1 \mathrm{~s}$ .

Re-arranging the formula to solve for $F_{avg}$ gives us:

F_{avg} = \frac{\Delta p}{\Delta t}.

Substituting the known values we have:

F_{avg} = \frac{3}{0.1} = 30 \mathrm{~N}.

The magnitude of the average force exerted by the hand of the player to catch the ball is $30 \mathrm{~N}$ .

Q40

Given below are two statements : Statement (I) : The limiting force of static friction depends on the area of contact and independent of materials. Statement (II) : The limiting force of kinetic friction is independent of the area of contact and depends on materials. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is incorrect but Statement II is correct

B Both Statement I and Statement II are correct

C Both Statement I and Statement II are incorrect

D Statement I is correct but Statement II is incorrect

Correct Answer

Option A

Solution

Let's analyze both statements: Statement (I): The limiting force of static friction depends on the area of contact and independent of materials.

This statement is incorrect.

The limiting force of static friction does not depend on the area of contact but is dependent on the materials in contact.

According to the law of static friction, the maximum static frictional force

f_s

that can occur before motion commences is given by the product of the coefficient of static friction

\mu_s

and the normal reaction force

N

:

f_s = \mu_s \times N

The coefficient

\mu_s

is a property that depends on the materials in contact, not on the area of contact. The normal force

N

is the force perpendicular to the surfaces in contact, influenced by the weight of the object and any other perpendicular forces acting upon it.

Statement (II): The limiting force of kinetic friction is independent of the area of contact and depends on materials.

This statement is correct.

Once an object is in motion, the kinetic frictional force

f_k

opposing its motion is given by the product of the coefficient of kinetic friction

\mu_k

and the normal force

N

:

f_k = \mu_k \times N

The coefficient

\mu_k

, like

\mu_s

, is also a property dependent on the materials of the surfaces in contact. It generally has a lower value than

\mu_s

, which is why objects tend to be easier to keep moving once they've started.

The kinetic frictional force is independent of the area of contact between the two surfaces.

Given these explanations, the correct answer would be: Option A: Statement I is incorrect but Statement II is correct.