Work Power & Energy — Page 9 | NEET Physics

Q81

A body of mass

2 \mathrm{~kg}

begins to move under the action of a time dependent force given by

\vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N

. The power developed by the force at the time

t

is given by:

A

\left(3 t^3+6 t^5\right) W

B

\left(9 t^5+6 t^3\right) W

C

\left(6 t^4+9 t^5\right) W

D

\left(9 t^3+6 t^5\right) W

Correct Answer

Option D

Solution

\begin{aligned} & \vec{F}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) N \\ & \vec{F}=m \vec{a}=\left(6 t \hat{i}+6 t^2 \hat{j}\right) \\ & \vec{a}=\frac{\vec{F}}{m}=\left(3 t \hat{i}+3 t^2 \hat{j}\right) \\ & \vec{v}=\int\limits_0^t \vec{a} d t=\frac{3 t^2}{2} \hat{i}+t^3 \hat{j} \\ & P=\vec{F} \cdot \vec{v}=\left(9 t^3+6 t^5\right) W \end{aligned}

Q82

A particle experiences a variable force

\overrightarrow F = \left( {4x\widehat i + 3{y^2}\widehat j} \right)

in a horizontal x-y plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the x-y plane, then Kinetic Energy changes by :

A 50.0 J

B 12.5 J

C 25.0 J

D 0 J

Correct Answer

Option C

Solution

W = \int {\overrightarrow F \,.\,d\overrightarrow r }

= \int\limits_1^2 {4xdx + \int\limits_2^3 {3{y^2}dy} }

= [2{x^2}]_1^2 + [{y^3}]_2^3

= 2 \times 3 + (27 - 8)

= 25

J

Q83

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement

x

is proportional to

A

x

B

{e^x}

C

{x^2}

D

{\log _e}x

Correct Answer

Option C

Solution

Given that, retardation $\propto$ displacement $\Rightarrow$

a=-kx

But we know

a = v{{dv} \over {dx}}\,\,\,\,\,\,\,\,\,

$\therefore$

{{vdv} \over {dx}} = - kx

\Rightarrow \int\limits_{{v_1}}^{{v_2}} v \,dv = - k\int\limits_0^x {xdx}

\left( {v_2^2 - v_1^2} \right) = - k{{{x^2}} \over 2}

\Rightarrow {1 \over 2}m\left( {v_2^2 - v_1^2} \right) = {1 \over 2}mk\left( {{{ - x^2} \over 2}} \right)

$\therefore$ Loss in kinetic energy is proportional to

{x^2}

. $\therefore$

\Delta K \propto {x^2}

Q84

An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being 68 kg, The mass of the elevator itself is 920 kg and it moves with a constant speed of 3 m/s. The frictional force opposing the motion is 6000 N. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator (g = 10 m/s2) must be at least :

A 48000 W

B 62360 W

C 56300 W

D 66000 W

Correct Answer

Option D

Solution

Net force on motor will be Fm = [920 + 68(10)]g + 6000 = 22000 N So, required power for motor P = Fm.V = 22000 $\times$ 3 = 66000 W

Q85

An object is dropped from a height h from the ground. Every time it hits the ground it looses 50% of its kinetic energy. The total distance covered as t

\to

\infty

is :

A 3h

B

\infty

C

{5 \over 3}

h

D

{8 \over 3}

h

Correct Answer

Option A

Solution

Let, Kinetic energy (k) =

{1 \over 2}

m

\upsilon

2 before it hit the ground. After hitting the ground kinetic energy (k') =

{1 \over 2}

m

\upsilon

_1^2

\therefore\,\,\,

According to the question,

{1 \over 2}

m

\upsilon

_1^2

=

{1 \over 2}

$\times$

{1 \over 2}

m

\upsilon

2 $\Rightarrow$

\,\,\,

\upsilon

1 =

{v \over {\sqrt 2 }}

After hitting the ground the object will bounce h' =

{{v_1^2} \over {2g}}

=

{{{v^2}} \over {4g}}

=

{h \over 2}

[ as h =

{{{v^2}} \over {2g}}

] Total distance travelled from the time it first hits the ground to the next time it hits the ground is =

{h \over 2}

+

{h \over 2}

= h So, this will create a infinite geometric progression with the common ration

{1 \over 2}

.

\therefore\,\,\,

Total distance covered = h (distance travelled by the obhect when first dropped, before it hits the ground) + (h +

{h \over 2}

+

{h \over 4}

+ . . . . . . . . $\propto$ ) = h +

{h \over {1 - {1 \over 2}}}

= h + 2h = 3h

Q86

The potential energy function (in

J

) of a particle in a region of space is given as

U=\left(2 x^2+3 y^3+2 z\right)

. Here

x, y

and

z

are in meter. The magnitude of

x

-component of force (in

N

) acting on the particle at point

P(1,2,3) \mathrm{m}

is :

A 4

B 2

C 8

D 6

Correct Answer

Option A

Solution

The force acting on a particle can be determined from the potential energy function by using the negative gradient.

In three-dimensional space, the force vector F is related to the potential energy function $U$ by:

\mathbf{F} = -\nabla U

where $\nabla U$ (the gradient of $U$ ) is a vector with components given by the partial derivatives of $U$ with respect to $x$ , $y$ , and $z$ :

\nabla U = \left( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right)

Given the potential energy function:

U = 2x^2 + 3y^3 + 2z

Calculate the partial derivative of $U$ with respect to $x$ to find the $x$ -component of the force: Partial Derivative with respect to $x$ :

\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x

Calculate the $x$ -component of the force: Since $F_x = -\dfrac{\partial U}{\partial x}$ , the $x$ -component of the force is:

F_x = -4x

Evaluate at point $P(1,2,3)$ : Substitute $x = 1$ into the expression for $F_x$ :

F_x = -4(1) = -4

The magnitude of the $x$ -component of the force is:

|F_x| = 4 \, \text{N}

Thus, the correct answer is Option A: 4.

Q87

The potential energy of a

1

kg

particle free to move along the

x

-axis is given by

V\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)J

. The total mechanical energy of the particle is

2J.

Then, the maximum speed (in

m/s

) is

A

{3 \over {\sqrt 2 }}

B

{\sqrt 2 }

C

{1 \over {\sqrt 2 }}

D

2

Correct Answer

Option A

Solution

Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum For minimum potential energy,

{{dV} \over {dx}} = 0

\Rightarrow {x^3} - x = 0

\Rightarrow x = \pm 1

$\Rightarrow$ Min. Potential energy (P.E.) =

{1 \over 4} - {1 \over 2} = - {1 \over 4}J

K.E{._{\left( {\max .} \right)}} + P.E{._{\left( {\min .} \right)}} = 2\,

(Given) $\therefore$

K.E{._{\left( {\max .} \right)}} = 2 + {1 \over 4} = {9 \over 4}

$\therefore$

{1 \over 2}mv_{\max }^2

=

{9 \over 4}

\Rightarrow {1 \over 2} \times 1 \times {v^2}_{\max .} = {9 \over 4}

\Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}

m/s

Q88

A mass of

M

kg

is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of

{45^ \circ }

with the initial vertical direction is

A

Mg\left( {\sqrt 2 + 1} \right)

B

Mg\sqrt 2

C

{{Mg} \over {\sqrt 2 }}

D

Mg\left( {\sqrt 2 - 1} \right)

Correct Answer

Option D

Solution

From work energy theorem we can say, Work done by tension $+$ work done by force (applied) $+$ Work done by gravitational force $=$ change in kinetic energy Here Work done by tension is zero

\Rightarrow 0 + F \times AB - Mg \times AC = 0

\Rightarrow F = Mg\left( {{{AC} \over {AB}}} \right) = Mg\left[ {{{1 - {1 \over {\sqrt 2 }}} \over {{1 \over 2}}}} \right]

[ as

AB = \ell \sin {45^ \circ } = {\ell \over {\sqrt 2 }}

and

AC = OC - OA = \ell - \ell \,\cos \,{45^ \circ } = \ell \left( {1 - {1 \over {\sqrt 2 }}} \right)

where

\ell =

length of the string. ]

\Rightarrow F = Mg\left( {\sqrt 2 - 1} \right)

Q89

Three bodies A, B and C have equal kinetic energies and their masses are

400 \mathrm{~g}, 1.2 \mathrm{~kg}

and

1.6 \mathrm{~kg}

respectively. The ratio of their linear momenta is :

A

1: \sqrt{3}: 2

B

\sqrt{3}: \sqrt{2}: 1

C

1: \sqrt{3}: \sqrt{2}

D

\sqrt{2} : \sqrt{3}: 1

Correct Answer

Option A

Solution

Given that the bodies A, B, and C have equal kinetic energies, we can use the relationship between kinetic energy (

K.E.

) and linear momentum (

p

) to find the ratio of their momenta. Recall the formula for kinetic energy is

K.E. = \frac{1}{2}mv^2

and the formula for momentum is

p = mv

, where

m

is the mass and

v

is the velocity of the object.

First, from the kinetic energy formula, we can express the velocity in terms of kinetic energy and mass:

v = \sqrt{\frac{2 \cdot K.E.}{m}}.

The momentum can then be rewritten using the velocity expression obtained from the kinetic energy equation:

p = m\sqrt{\frac{2 \cdot K.E.}{m}} = \sqrt{2m \cdot K.E.}.

Given that the kinetic energies are the same for all three bodies, we can ignore the kinetic energy term when comparing the ratios, simplifying our comparison to the square root of their masses:

p \propto \sqrt{m}.

Now, we calculate the ratio of their linear momenta using their masses.

Note that the masses should be in consistent units for a valid comparison, so we'll use kilograms for all: Mass of A =

400 \mathrm{~g} = 0.4 \mathrm{~kg}

Mass of B =

1.2 \mathrm{~kg}

Mass of C =

1.6 \mathrm{~kg}

Thus, the ratio of their momenta will be proportional to the square root of their masses:

\text{Ratio of momenta} = \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6} = \sqrt{\frac{4}{10}} : \sqrt{\frac{12}{10}} : \sqrt{\frac{16}{10}} = \sqrt{\frac{2}{5}} : \sqrt{\frac{6}{5}} : \sqrt{\frac{8}{5}}.

Simplifying these we get:

\text{Ratio of momenta} = \frac{\sqrt{2}}{\sqrt{5}} : \frac{\sqrt{6}}{\sqrt{5}} : \frac{\sqrt{8}}{\sqrt{5}} = \sqrt{2} : \sqrt{6} : \sqrt{8}.

Recognizing that

\sqrt{6}

is equivalent to

\sqrt{2} \cdot \sqrt{3}

and that

\sqrt{8}

is equivalent to

\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2} = 2\sqrt{2}

, we see this can also be expressed as:

\sqrt{2} : \sqrt{2} \cdot \sqrt{3} : 2\sqrt{2}.

Dividing through by

\sqrt{2}

to simplify the ratio, the final ratio of their linear momenta is:

1 : \sqrt{3} : 2,

which matches Option A

1: \sqrt{3}: 2.

Q90

Identify the correct statements from the following : A. Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket is negative. B. Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative. C. Work done by friction on a body sliding down an inclined plane is positive. D. Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero. E. Work done by the air resistance on an oscillating pendulum is negative. Choose the correct answer from the options given below :

A A and C only

B B and D only

C B, D and E only

D B and E only

Correct Answer

Option D

Solution

When a man lifts a bucket out of a well using a rope, work is done by the man and the gravitational force.

The work done by the man is positive as he has to exert an upward force to lift the bucket.

The work done by the gravitational force is negative because the direction of the force is opposite to the direction of displacement.

Therefore, the statement (A) "Work done by a man in lifting a bucket out of a well by means of rope tied to the bucket is negative." is incorrect.

Therefore, the statement (B) "Work done by gravitational force in lifting a bucket out of a well by a rope tied to the bucket is negative." is correct.

Work is defined as the product of force and displacement in the direction of the force.

When a body slides down an inclined plane, the force of friction acts against the motion of the body, opposing its descent.

The direction of the force of friction is opposite to the direction of the displacement of the body, which is downwards.

Hence, the work done by the force of friction is negative.

Therefore, the statement (C) "Work done by friction on a body sliding down an inclined plane is positive" is incorrect.

If the body is moving on a rough horizontal plane, there will be friction present, which will act in the opposite direction to the applied force.

The force of friction will oppose the motion of the body, reducing its velocity.

As a result, the net work done on the body will not be zero, as the force of friction and the applied force will not cancel each other out completely.

Therefore, the statement (D) "Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity is zero." is incorrect.

Statement E: "Work done by the air resistance on an oscillating pendulum is negative."

This statement refers to the work done by the air resistance on an oscillating pendulum, which is a physical system that swings back and forth under the influence of gravity.

As the pendulum oscillates, it experiences air resistance, which opposes its motion and slows it down.

The direction of the air resistance force is opposite to the direction of the displacement of the pendulum, which is back and forth.

Hence, the work done by the air resistance force is negative, as the direction of the force and the displacement are opposite.

Therefore, the statement (E) "Work done by the air resistance on an oscillating pendulum is negative" is correct.