Laws of Motion — Page 6 | JEE Physics

Q51

An object with mass 500 g moves along x-axis with speed

v = 4\sqrt{x}

m/s. The force acting on the object is :

A 8 N

B 4 N

C 5 N

D 6 N

Correct Answer

Option B

Solution

To find the force acting on the object, we use the formula $F = m \times a$ , where $F$ is the force, $m$ is the mass, and $a$ is the acceleration.

The velocity $v$ is given as $v = 4 \sqrt{x}$ .

Squaring both sides, we have $v^2 = 16x$ .

Differentiate $v^2$ with respect to $x$ : $2v \dfrac{dv}{dx} = 16$ Solving for $\dfrac{dv}{dx}$ , we find: $\dfrac{v \, dv}{dx} = \dfrac{16}{2} = 8$ The force $F$ is then calculated as: $F = 0.5 \times 8 = 4 \, \text{N}$ Thus, the force acting on the object is 4 N.

Q52

A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle

\theta

with the horizontal, a point object of mass m is kept. The minimum coefficient of friction

\mu

min between the mass and the inclined surface such that the mass does not move is :

A tan

\theta

B 2tan

\theta

C 3tan

\theta

D tan2

\theta

Correct Answer

Option A

Solution

Rocket is moving upward with acceleration 2g and gravitation acceleration is g downward direction.

So, acceleration experienced by the point object, = 2g $-$ ( $-$ g) = 3g At equilibrium, N = 3mgcos $\theta$ $\mu$ N = 3mgsin $\theta$ $\Rightarrow$ $\mu$ (3mgcos $\theta$ ) = 3mg sin $\theta$ $\Rightarrow$ $\mu$ = tan $\theta$

Q53

A smooth block is released at rest on a

{45^ \circ }

incline and then slides a distance

'd'

. The time taken to slide is

'n'

times as much to slide on rough incline than on a smooth incline. The coefficient of friction is

A

{\mu _k} = \sqrt {1 - {1 \over {{n^2}}}}

B

{\mu _k} = 1 - {1 \over {{n^2}}}

C

{\mu _k} = \sqrt {1 - {1 \over {{n^2}}}}

D

{\mu _s} = 1 - {1 \over {{n^2}}}

Correct Answer

Option B

Solution

For smooth surface,

d = {1 \over 2}\left( {g\,\sin \,\theta } \right)t_1^2,

{t_1} = \sqrt {{{2d} \over {g\,\sin \,\theta }}} ,

When surface is rough

d = {1 \over 2}\left( {g\,\sin \,\theta - \mu g\,\cos \theta } \right)t_2^2

{t_2} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}}

According to question,

{t_2} = n{t_1}

n\sqrt {{{2d} \over {g\,\sin \,\theta }}} = \sqrt {{{2d} \over {g\,\sin \,\theta - \mu g\,\cos \theta }}}

n = {1 \over {\sqrt {1 - {\mu _k}} }}

( as

\cos \,{45^ \circ } = \sin \,{45^ \circ } = {1 \over {\sqrt 2 }}

)

{n^2} = {1 \over {1 - {\mu _k}}}

or

1 - {\mu _k} = {1 \over {{n^2}}}

or

{\mu _k} = 1 - {1 \over {{n^2}}}

Q54

Three forces

F_{1}=10 \mathrm{~N}, F_{2}=8 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}

are acting on a particle of mass

5 \mathrm{~kg}

. The forces

\mathrm{F}_{2}

and

\mathrm{F}_{3}

are applied perpendicularly so that particle remains at rest. If the force

F_{1}

is removed, then the acceleration of the particle is:

A

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

B

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

C

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

D

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

Correct Answer

Option C

Solution

Since the particle is initially at rest, the net force acting on it is zero. This means that the forces

F_1

,

F_2

, and

F_3

are balanced. Given that

F_2

and

F_3

are acting perpendicularly, we can represent the balance of forces as follows:

F_1 = \sqrt{F_2^2 + F_3^2}

We can now calculate the equivalent force resulting from

F_2

and

F_3

:

F_1 = \sqrt{(8 \mathrm{~N})^2 + (6 \mathrm{~N})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \mathrm{~N}

Since the particle is initially at rest, when the force

F_1

is removed, only

F_2

and

F_3

remain. The net force acting on the particle is the equivalent force resulting from

F_2

and

F_3

, which we have calculated to be

10 \mathrm{~N}

. Now, we can use Newton's second law to find the acceleration of the particle:

F = ma

Where:

F

is the net force acting on the particle

m

is the mass of the particle

a

is the acceleration of the particle Rearranging the equation to solve for

a

:

a = \frac{F}{m}

Plugging in the values for

F

and

m

:

a = \frac{10 \mathrm{~N}}{5 \mathrm{~kg}} = 2 \mathrm{~ms}^{-2}

The acceleration of the particle when the force

F_1

is removed is

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

, which corresponds to Option C.

Q55

The initial mass of a rocket is 1000 kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of 20 ms-2. The gases come out at a relative speed of 500 ms

-

1 with respect to the rocket : [Use g = 10 m/s2]

A 6.0

\times

102 kg s

-

1

B 500 kg s

-

1

C 10 kg s

-

1

D 60 kg s

-

1

Correct Answer

Option D

Solution

{F_{thrust}} = \left( {{{dm} \over {dt}}.{V_{rel}}} \right)

\left( {{{dm} \over {dt}}{V_{rel}} - mg} \right) = ma

\Rightarrow \left( {{{dm} \over {dt}}} \right) \times 500 - {10^3} \times 10 = {10^3} \times 20

{{dm} \over {dt}}

= (60 kg /s)

Q56

A heavy iron bar, of weight

W

is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle

\theta

with the horizontal. The weight experienced by the person is :

A

W \sin \theta

B

W

C

\dfrac{W}{2}

D

W \cos \theta

Correct Answer

Option C

Solution

\begin{aligned} & m g \times \frac{L}{2} \cos \theta=N \times L \cos \theta \\ & \Rightarrow \quad N=\frac{m g}{2}=\frac{W}{2} \end{aligned}

Q57

A marble block of mass

2

kg

lying on ice when given a velocity of

6

m/s

is stopped by friction in

10

s.

Then the coefficient of friction is

A

0.02

B

0.03

C

0.04

D

0.06

Correct Answer

Option D

Solution

The retarding force is created by the frictional force. Given,

u = 6m/s,\,v = 0,\,t = 10s,

Retardation

(a) = - {f \over m} = {{ - umg} \over m} = - \mu g = - 10\mu

v = u + at

0 = 6 - 10\mu \times 10

$\therefore$

\mu = 0.06

Q58

The position vector of a particle related to time

t

is given by

\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m

The direction of net force experienced by the particle is :

A Positive

x

- axis

B Positive

y

- axis

C Positive

z

- axis

D In

x

-

y

plane

Correct Answer

Option B

Solution

To find the direction of the net force experienced by the particle, we need to find the acceleration vector of the particle and then use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration vector.

The position vector of the particle is given by:

\vec{r} = (10t\hat{i} + 15t^2\hat{j} + 7\hat{k})\,\text{m}

Differentiating $\vec{r}$ twice with respect to time $t$ , we get the acceleration vector:

\vec{a} = \frac{d^2\vec{r}}{dt^2} = \frac{d}{dt}(10\hat{i} + 30t\hat{j}) = 30\hat{j}\,\text{m/s}^2

Therefore, the acceleration vector is $\vec{a} = 30\hat{j}\,\text{m/s}^2$ .

Using Newton's second law, the net force on the particle is given by:

\vec{F}_{net} = m\vec{a}

where $m$ is the mass of the particle.

Since we are only interested in the direction of the net force, we can ignore the magnitude of the acceleration and focus on its direction, which is along the positive $y$ -axis.

Q59

A lift is moving down with acceleration

a.

A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively

A

g,g

B

g-a, g-a

C

g-a, g

D

a, g

Correct Answer

Option C

Solution

Let acceleration of ball =

{\overrightarrow a _b}

and acceleration of man is =

{\overrightarrow a _m}

With respect to the man standing in the lift, the acceleration of the ball

{\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}

\Rightarrow {a_{bm}} = g - a

Where

a

is the acceleration of the man as the acceleration of the lift is

a

. With respect to the man standing on the ground the acceleration of the ball

{\overrightarrow a _{bm}} = {\overrightarrow a _b} - {\overrightarrow a _m}

\Rightarrow {a_{bm}} = g - 0

= g

Q60

Given below are two statements : Statement I : An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable. Statement II : Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Both Statement I and Statement II are true

C Statement I is false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option D

Solution

Statement I says that when the weight of an elevator is balanced with the tension of its cable, it can move up or down with a uniform speed.

This is true because the weight of the elevator is balanced by the tension in the cable, which allows it to move smoothly and at a constant speed.

Statement II says that the force exerted by the floor of an elevator on a person's foot is greater than their weight when the elevator goes down with increasing speed.

This is false because the force exerted by the floor on a person's foot is equal to their weight, regardless of the speed of the elevator.

The person's weight is a constant force and does not change with the speed of the elevator.

The apparent weight of a person may change with the speed of the elevator, but this is due to the effects of acceleration and not an increase in the force exerted by the floor.