Laws of Motion — Page 5 | JEE Physics

Q41

A block of mass

m

is placed on a surface having vertical crossection given by

y=x^2 / 4

. If coefficient of friction is 0.5, the maximum height above the ground at which block can be placed without slipping is:

A 1/2 m

B 1/3 m

C 1/6 m

D 1/4 m

Correct Answer

Option D

Solution

Given: The equation of the surface:

y = \frac{x^2}{4}

. Coefficient of friction:

\mu = 0.5

. Gravitational acceleration:

g

(assumed constant). 1. Slope of the Surface: The slope of the surface at any point is given by:

\tan\theta = \frac{dy}{dx}

From the equation of the surface:

y = \frac{x^2}{4}

Differentiating with respect to $x$ :

\frac{dy}{dx} = \frac{x}{2}

Thus, the slope at any point is:

\tan\theta = \frac{x}{2}

2.

Forces Acting on the Block: At the point where the block is placed: Weight of the block acts vertically downward: $mg$ .

Normal force acts perpendicular to the surface.

Frictional force acts parallel to the surface, opposing the component of the weight that causes slipping.

3.

Condition for No Slipping: The block will not slip if the frictional force is sufficient to counteract the component of the gravitational force parallel to the slope.

The frictional force is:

f = \mu N

The normal force $N$ is given by:

N = mg \cos\theta

The component of weight parallel to the slope is:

F_{\text{parallel}} = mg \sin\theta

For the block to not slip:

f \geq F_{\text{parallel}}

Substitute $f = \mu N$ and $N = mg \cos\theta$ :

\mu (mg \cos\theta) \geq mg \sin\theta

Simplify:

\mu \cos\theta \geq \sin\theta

Divide through by $\cos\theta$ :

\mu \geq \tan\theta

4. Maximum Slope Without Slipping: From the above condition:

\tan\theta \leq \mu

Substitute $\mu = 0.5$ :

\tan\theta \leq 0.5

Thus:

\frac{x}{2} \leq 0.5

x \leq 1

5. Maximum Height: The height $y$ of the block at $x = 1$ is obtained from the surface equation:

y = \frac{x^2}{4}

Substitute $x = 1$ :

y = \frac{1^2}{4} = \frac{1}{4} \, \text{m}

Final Answer: The maximum height above the ground at which the block can be placed without slipping is:

\boxed{\frac{1}{4} \, \text{m}}

Thus, the correct option is Option D.

Q42

A light unstretchable string passing over a smooth light pulley connects two blocks of masses

m_1

and

m_2

. If the acceleration of the system is

\dfrac{g}{8}

, then the ratio of the masses

\dfrac{m_2}{m_1}

is :

A

5: 3

B

8: 1

C

9: 7

D

4: 3

Correct Answer

Option C

Solution

To find the ratio of the masses

\frac{m_2}{m_1}

given that the acceleration of the system is

\frac{g}{8}

, where

g

is the acceleration due to gravity, we need to analyze the forces acting on each mass and apply Newton's second law of motion.

For mass

m_1

, the force acting downward (towards the centre of the Earth) is the gravitational force

m_1g

, and the tension

T

acts upward. Its equation of motion, considering downward as positive, can be given by: $m_1g - T = m_1a$ For mass

m_2

, the force acting downward is

m_2g

, but since it is on the opposite side of the pulley, the tension

T

is upwards (considering upward movement as positive direction), so we have: $T - m_2g = m_2a$ Since the pulley is light and smooth, the tension

T

is the same on both sides of the pulley. Also, the system accelerates together, so

a = \frac{g}{8}

. Adding the two equations to eliminate

T

gives us: $m_1g - m_2g = (m_1 - m_2)a$ $g(m_1 - m_2) = (m_1 - m_2)\dfrac{g}{8}$ Canceling out

g

and dividing both sides by

m_1 - m_2

(assuming

m_1 \neq m_2

), we get: $1 = \dfrac{1}{8}$ This simplification doesn't align with finding the ratio directly, indicating a mistake in handling the simultaneous equations relation to

a

and

T

.

Let's correct our approach to finding the ratio of the masses based on the acceleration and tension.

Since the acceleration

a = \frac{g}{8}

, and considering the system as a whole, the net force causing the acceleration is the difference in gravitational forces on the two masses.

This net force provides the system's entire acceleration.

Thus, correctly setting up the equations for the system should give: $m_2g - m_1g = (m_1 + m_2)\dfrac{g}{8}$ $g(m_2-m_1) = \dfrac{g}{8}(m_1 + m_2)$ $8(m_2 - m_1) = m_1 + m_2$ $8m_2 - 8m_1 = m_1 + m_2$ $7m_2 = 9m_1$ $\dfrac{m_2}{m_1} = \dfrac{9}{7}$ Therefore, the correct ratio of the masses

\frac{m_2}{m_1}

is

9:7

, which corresponds to Option C.

Q43

A given object takes

\mathrm{n}

times the time to slide down

45^{\circ}

rough inclined plane as it takes the time to slide down an identical perfectly smooth

45^{\circ}

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

A

1-\dfrac{1}{\mathrm{n}^2}

B

\sqrt{1-\dfrac{1}{\mathrm{n}^2}}

C

1-n^2

D

\sqrt{1-n^2}

Correct Answer

Option A

Solution

To determine the coefficient of kinetic friction, let's analyze the time taken by the object to slide down each plane and use the equations of motion for both scenarios.

First, consider the perfectly smooth

45^{\circ}

inclined plane (no friction).

The acceleration of the object on this plane can be calculated using the component of gravitational force parallel to the incline.

Since there is no friction, the only force acting down the plane is the component of the gravitational force:

a_{\text{smooth}} = g \sin 45^{\circ} = \frac{g}{\sqrt{2}}

Let the time taken to slide down this smooth plane be

t_{\text{smooth}}

. The distance

d

covered by the object can be expressed using the equation of motion:

d = \frac{1}{2} a_{\text{smooth}} t_{\text{smooth}}^2 = \frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2

Now, consider the rough

45^{\circ}

inclined plane.

The acceleration down the rough plane can be found by considering both the component of gravitational force and the kinetic friction force.

Here, the friction force is

f_k = \mu_k N

, where

N = mg \cos 45^\circ = \frac{mg}{\sqrt{2}}

. Thus the frictional force is:

f_k = \mu_k \cdot \frac{mg}{\sqrt{2}}

The net force acting on the object down the plane would be:

F_{\text{net}} = mg \sin 45^{\circ} - f_k = \frac{mg}{\sqrt{2}} - \mu_k \cdot \frac{mg}{\sqrt{2}}

So the net acceleration

a_{\text{rough}}

is:

a_{\text{rough}} = \frac{F_{\text{net}}}{m} = \frac{g}{\sqrt{2}} - \mu_k \cdot \frac{g}{\sqrt{2}} = \frac{g}{\sqrt{2}} (1 - \mu_k)

The time taken to slide down the rough plane is given as

nt_{\text{smooth}}

. So, the distance

d

can be written as:

d = \frac{1}{2} a_{\text{rough}} (nt_{\text{smooth}})^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2

Setting the distances equal for both cases, we get:

\frac{1}{2} \frac{g}{\sqrt{2}} t_{\text{smooth}}^2 = \frac{1}{2} \left(\frac{g}{\sqrt{2}} (1 - \mu_k)\right) (nt_{\text{smooth}})^2

Simplifying, we find:

t_{\text{smooth}}^2 = (1 - \mu_k) n^2 t_{\text{smooth}}^2

Therefore:

1 = (1 - \mu_k) n^2

Solving for

\mu_k

, we get:

\mu_k = 1 - \frac{1}{n^2}

Thus, the coefficient of kinetic friction between the object and the surface of inclined plane is: Option A:

1 - \frac{1}{\mathrm{n}^2}

Q44

A player caught a cricket ball of mass

150 \mathrm{~g}

moving at a speed of

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

. If the catching process is completed in

0.1 \mathrm{~s}

, the magnitude of force exerted by the ball on the hand of the player is:

A 150 N

B 3 N

C 30 N

D 300 N

Correct Answer

Option C

Solution

The force exerted by the ball on the hand can be calculated using the formula derived from Newton's second law of motion, which is

F = \frac{\Delta p}{\Delta t}

, where

F

is the force,

\Delta p

represents the change in momentum, and

\Delta t

is the time over which this change occurs. The change in momentum,

\Delta p

, can be calculated as the difference between the final momentum,

p_f

, and the initial momentum,

p_i

.

In this scenario, because the ball comes to a stop in the player's hand, its final velocity (and hence, its final momentum) is 0.

Therefore, the change in momentum is equal to the initial momentum of the ball (since final momentum is zero).

The initial momentum,

p_i

, of the ball can be calculated using the formula

p = mv

, where

m

is the mass of the ball and

v

is its velocity. Given that the mass of the ball is

150 \, \mathrm{g} = 0.15 \, \mathrm{kg}

(converting grams to kilograms) and its velocity is

20 \, \mathrm{m/s}

, we have:

p_i = (0.15 \, \mathrm{kg}) \times (20 \, \mathrm{m/s}) = 3 \, \mathrm{kg \cdot m/s}

Since the change in momentum,

\Delta p

, equals the initial momentum (

p_i

) because the final momentum is 0, the force exerted can be found by substituting

\Delta p

and

\Delta t

into the first formula:

F = \frac{3 \, \mathrm{kg \cdot m/s}}{0.1 \, \mathrm{s}} = 30 \, \mathrm{N}

Therefore, the magnitude of force exerted by the ball on the hand of the player is 30 N, which corresponds to Option C.

Q45

A wooden block of mass

5 \mathrm{~kg}

rests on a soft horizontal floor. When an iron cylinder of mass

25 \mathrm{~kg}

is placed on the top of the block, the floor yields and the block and the cylinder together go down with an acceleration of

0.1 \mathrm{~ms}^{-2}

. The action force of the system on the floor is equal to :

A 297 N

B 291 N

C 196 N

D 294 N

Correct Answer

Option B

Solution

To solve this problem, we'll analyze the forces acting on the combined system of the wooden block and the iron cylinder as they accelerate downward due to the yielding floor.

Given: Mass of wooden block, $m_1 = 5 \, \text{kg}$ Mass of iron cylinder, $m_2 = 25 \, \text{kg}$ Total mass of the system, $m = m_1 + m_2 = 30 \, \text{kg}$ Acceleration of the system, $a = 0.1 \, \text{m/s}^2$ (downward) Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$ Step 1: Draw the Free Body Diagram (FBD) For the combined system: Downward Forces: Weight of the system: $W = m \times g = 30 \times 9.8 = 294 \, \text{N}$ Upward Forces: Normal force from the floor: $N$ Step 2: Apply Newton's Second Law Since the system is accelerating downward, the net force is: $\text{Net Force} = \text{Total Downward Force} - \text{Total Upward Force}$ According to Newton's second law: $m \times a = W - N$ Step 3: Solve for the Normal Force ( $N$ ) $N = W - m \times a$ Substitute the given values: $N = 294 \, \text{N} - 30 \, \text{kg} \times 0.1 \, \text{m/s}^2$ $N = 294 \, \text{N} - 3 \, \text{N}$ $N = 291 \, \text{N}$ Step 4: Interpret the Result The normal force ( $N$ ) is the force exerted by the floor on the system upward.

By Newton's third law, the action force of the system on the floor is equal in magnitude and opposite in direction to the normal force.

Therefore, the action force of the system on the floor is $291 \, \text{N}$ .

Q46

A heavy box of mass

50 \mathrm{~kg}

is moving on a horizontal surface. If co-efficient of kinetic friction between the box and horizontal surface is 0.3 then force of kinetic friction is :

A 1470 N

B 147 N

C 1.47 N

D 14.7 N

Correct Answer

Option B

Solution

To find the force of kinetic friction acting on the box, we can use the formula for kinetic friction, which is given by: $F_{\text{friction}} = \mu_{k} \cdot N$ where: $F_{\text{friction}}$ is the force of friction, $\mu_{k}$ is the coefficient of kinetic friction, $N$ is the normal force exerted by the surface onto the box.

Since the box is moving on a horizontal surface, the normal force $N$ would be equal to the weight of the box, which is calculated by $mg$ , where $m$ is the mass of the box and $g$ is the acceleration due to gravity.

For most calculations, $g$ is approximated as

9.8 \, \text{m/s}^2

.

Substituting the given values: $m = 50 \, \text{kg}$ , $\mu_k = 0.3$ , $g = 9.8 \, \text{m/s}^2$ .

First, calculate the weight of the box, which is the normal force: $N = mg = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 490 \, \text{N}$ Then, use this value to find the force of kinetic friction: $F_{\text{friction}} = \mu_{k} \cdot N = 0.3 \times 490 \, \text{N} = 147 \, \text{N}$ Thus, the force of kinetic friction acting on the box is 147 N.

The correct answer is Option B.

Q47

A body of weight

200 \mathrm{~N}

is suspended from a tree branch through a chain of mass

10 \mathrm{~kg}

. The branch pulls the chain by a force equal to (if

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

) :

A 300 N

B 100 N

C 150 N

D 200 N

Correct Answer

Option A

Solution

To determine the force that the branch pulls on the chain, we need to consider the combined weight of the body and the chain, as this is the total force that the branch must support due to gravity.

The weight of the body is given as

200\, \mathrm{N}

.

To find the weight of the chain, we use the formula for weight, which is the mass of an object multiplied by the acceleration due to gravity (

g

). The mass of the chain is given as

10\, \mathrm{kg}

and

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

, so:

\text{Weight of the chain} = \text{Mass of the chain} \times g

= 10\, \mathrm{kg} \times 10\, \mathrm{m/s}^2

= 100\, \mathrm{N}

The total force that the branch must support is the sum of the weight of the body and the weight of the chain:

\text{Total force} = \text{Weight of the body} + \text{Weight of the chain}

= 200\, \mathrm{N} + 100\, \mathrm{N}

= 300\, \mathrm{N}

Therefore, the branch pulls the chain by a force of

300\, \mathrm{N}

, which corresponds to Option A.

Q48

A balloon and its content having mass M is moving up with an acceleration ‘a’. The mass that must be released from the content so that the balloon starts moving up with an acceleration ‘3a’ will be(Take ‘g’ as acceleration due to gravity)

A

\dfrac{3Ma}{2a + g}

B

\dfrac{2Ma}{3a + g}

C

\dfrac{3Ma}{2a - g}

D

\dfrac{2Ma}{3a - g}

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & F-m g=m a \\ & F=m a+m g \\ & F-(m-x) g=(m-x) 3 a \end{aligned}\\ &\text{ Put F }\\ &\begin{aligned} & \mathrm{Ma}+\mathrm{mg}-\mathrm{mg}+\mathrm{xg}=3 \mathrm{ma}-3 \mathrm{xa} \\ & \mathrm{x}=\frac{2 \mathrm{ma}}{\mathrm{~g}+3 \mathrm{a}} \end{aligned} \end{aligned}

Q49

A body of mass 2 kg moving with velocity of

\vec{v}_{in} = 3 \hat{i} + 4 \hat{j} \text{ ms}^{-1}

enters into a constant force field of 6N directed along positive z-axis. If the body remains in the field for a period of

\dfrac{5}{3}

seconds, then velocity of the body when it emerges from force field is.

A

3\hat{i} + 4\hat{j} + \sqrt{5} \hat{k}

B

4\hat{i} + 3\hat{j} + 5\hat{k}

C

3\hat{i} + 4\hat{j} - 5\hat{k}

D

3\hat{i} + 4\hat{j} + 5\hat{k}

Correct Answer

Option D

Solution

To determine the velocity of the body as it emerges from the force field, we can break this process down as follows: Force and Acceleration Calculation: The body is subjected to a constant force of 6 N, which acts along the positive z-axis.

The mass of the body is 2 kg.

Using Newton's second law, $\vec{F} = m \vec{a}$ , the acceleration $\vec{a}$ can be calculated as: $\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{6 \hat{k}}{2} = 3 \hat{k} \text{ ms}^{-2}$ Initial Velocity: The initial velocity of the body is given by: $\vec{u} = 3 \hat{i} + 4 \hat{j} \text{ ms}^{-1}$ Time of Motion: The time the body spends in the force field is: $t = \dfrac{5}{3} \text{ seconds}$ Final Velocity Calculation: The final velocity $\vec{v}$ is given by the equation of motion: $\vec{v} = \vec{u} + \vec{a} t$ Substituting the known values: $\vec{v} = (3 \hat{i} + 4 \hat{j}) + 3 \hat{k} \left(\dfrac{5}{3}\right)$ Simplifying the expression gives: $\vec{v} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}$ Therefore, the velocity of the body as it exits the force field is $3 \hat{i} + 4 \hat{j} + 5 \hat{k} \text{ ms}^{-1}$ .

Q50

A cubic block of mass

m

is sliding down on an inclined plane at

60^{\circ}

with an acceleration of

\dfrac{g}{2}

, the value of coefficient of kinetic friction is

A

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

B

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

C

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

D

\sqrt{3}-1

Correct Answer

Option D

Solution

\begin{aligned} & m g \sin 60^{\circ}-\mu m g \cos 60^{\circ}=m a \\ & g \sin 60-\mu g \cos 60=\frac{g}{2} \\ & \frac{\sqrt{3}}{2}-\frac{\mu}{2}=\frac{1}{2} \\ & \mu=\sqrt{3}-1 \end{aligned}