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Work Power & Energy

NEET Physics · 97 questions · Page 1 of 10 · Click an option or "Show Solution" to reveal answer

Q1
The power of a crane, which lifts a mass of 1000 kg to a height of 20 m in 10 s is: (g=9.8 m/s2)\left(g=9.8 \mathrm{~m} / \mathrm{s}^2\right)
A 19.6 W
B 39.2 W
C 19.6 kW
D 39.2 kW
Correct Answer
Option C
Solution
 Power = Work  Time =mght=103×9.8×2010=19.6 kW\begin{aligned} & \text{ Power }=\frac{\text{ Work }}{\text{ Time }} \\ & =\frac{m g h}{t} \\ & =\frac{10^3 \times 9.8 \times 20}{10}=19.6 \mathrm{~kW} \end{aligned}
Q2
The kinetic energies of two similar cars AA and BB are 100 J and 225 J respectively. On applying breaks, car AA stops after 1000 m and car BB stops after 1500 m . If FAF_A and FBF_B are the forces applied by the breaks on cars AA and BB respectively, then the ratio of FAFB\dfrac{F_A}{F_B} is
A 13\dfrac{1}{3}
B 12\dfrac{1}{2}
C 32\dfrac{3}{2}
D 23\dfrac{2}{3}
Correct Answer
Option D
Solution

According to the work-energy theorem, the work done by the braking force is equal to the change in kinetic energy.

Thus, for each car, we have: FS=ΔKE FS = \Delta KE This can be rearranged to: −FS=kf−ki⇒FS=ki−kf -FS = k_f - k_i \quad \Rightarrow \quad FS = k_i - k_f For both cars, since the final kinetic energy kf k_f is zero when they stop, the equation simplifies to: FS=ki FS = k_i For car A A and car B B , the forces FA F_A and FB F_B applied by the brakes satisfy: FA⋅SA=kAandFB⋅SB=kB F_A \cdot S_A = k_A \quad \text{and} \quad F_B \cdot S_B = k_B Solving for the ratio of the forces: FAFB=kAkB×SBSA \dfrac{F_A}{F_B} = \dfrac{k_A}{k_B} \times \dfrac{S_B}{S_A} Substituting the given kinetic energies and stopping distances: FAFB=100225×15001000 \dfrac{F_A}{F_B} = \dfrac{100}{225} \times \dfrac{1500}{1000} This simplifies to: =150225=23 = \dfrac{150}{225} = \dfrac{2}{3} Thus, the ratio of the forces FAFB \dfrac{F_A}{F_B} is 23 \dfrac{2}{3} .

Q3
A bob of heavy mass mm is suspended by a light string of length //. The bob is given a horizontal velocity v0v_0 as shown in figure. If the string gets slack at some point PP making an angle θ\theta from the horizontal, the ratio of the speed vv of the bob at point PP to its initial speed v0v_0 is:
A (cos⁡θ2+3sin⁡θ)12\left(\dfrac{\cos \theta}{2+3 \sin \theta}\right)^{\dfrac{1}{2}}
B (sin⁡θ2+3sin⁡θ)12\left(\dfrac{\sin \theta}{2+3 \sin \theta}\right)^{\dfrac{1}{2}}
C (sin⁡θ)12(\sin \theta)^{\dfrac{1}{2}}
D (12+3sin⁡θ)12\left(\dfrac{1}{2+3 \sin \theta}\right)^{\dfrac{1}{2}}
Correct Answer
Option B
Solution

At Point P,mgsin⁡θ=mv2l..... (1)P, m g \sin \theta=\dfrac{m v^2}{l}\quad\text{..... (1)} By conservation of mechanical energy at point P∈QP \in Q

12mv02=12mv2+mg(I+Isin⁡θ)\frac{1}{2} m v_0^2=\frac{1}{2} m v^2+m g(I+I \sin \theta)
v022=v22+gl(1+sin⁡θ) Put gl =v2sin⁡θ using (1) v022=v22+v2sin⁡θ(1+sin⁡θ)v022=v22+v2sin⁡θ+v2v022=32v2+2v22sin⁡θv02=v2[3+2sin⁡θ]vv0=(sin⁡θ3sin⁡θ+2)12\begin{aligned} & \frac{v_0^2}{2}=\frac{v^2}{2}+g l(1+\sin \theta) \\ & \text{ Put gl }=\frac{v^2}{\sin \theta} \text{ using (1) } \\ & \frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta}(1+\sin \theta) \\ & \frac{v_0^2}{2}=\frac{v^2}{2}+\frac{v^2}{\sin \theta}+v^2 \\ & \frac{v_0^2}{2}=\frac{3}{2} v^2+\frac{2 v^2}{2 \sin \theta} \\ & v_0^2=v^2\left[3+\frac{2}{\sin \theta}\right] \\ & \frac{v}{v_0}=\left(\frac{\sin \theta}{3 \sin \theta+2}\right)^{\frac{1}{2}} \end{aligned}
Q4
An object moving along horizontal xx-direction with kinetic energy 10 J10 \mathrm{~J} is displaced through x=(3i^)mx=(3 \hat{i}) \mathrm{m} by the force F⃗=(−2i^+3j^)N\vec{F}=(-2 \hat{i}+3 \hat{j}) \mathrm{N}. The kinetic energy of the object at the end of the displacement xx is
A 10 J10 \mathrm{~J}
B 16 J16 \mathrm{~J}
C 4 J4 \mathrm{~J}
D 6 J6 \mathrm{~J}
Correct Answer
Option C
Solution

To find the kinetic energy of the object at the end of the displacement, we need to calculate the work done by the force

F⃗\vec{F}

on the object during its displacement.

The work done by a force is given by the dot product of the force and the displacement vectors:

W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d}

Here, the force

F⃗\vec{F}

is given as:

F⃗=−2i^+3j^ N\vec{F} = -2 \hat{i} + 3 \hat{j} \, \mathrm{N}

And the displacement

d⃗\vec{d}

is given as:

d⃗=3i^ m\vec{d} = 3 \hat{i} \, \mathrm{m}

The dot product of the two vectors is calculated as follows:

W=(−2i^+3j^)⋅(3i^)W = (-2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i})

Since the

j^\hat{j}

component of the force does not contribute to the work done in the

xx

-direction, it can be ignored. Thus, the dot product yields:

W=−2⋅3 N⋅m=−6 JW = -2 \cdot 3 \, \mathrm{N} \cdot \mathrm{m} = -6 \, \mathrm{J}

The negative sign indicates that the force does work against the direction of displacement.

Now, the initial kinetic energy of the object is:

Ki=10 JK_i = 10 \, \mathrm{J}

The work-energy theorem relates the work done on an object to its change in kinetic energy:

W=Kf−KiW = K_f - K_i

Rearranging for the final kinetic energy

KfK_f

gives:

Kf=Ki+WK_f = K_i + W

Substituting the known values:

Kf=10 J+(−6 J)K_f = 10 \, \mathrm{J} + (-6 \, \mathrm{J})

This simplifies to:

Kf=4 JK_f = 4 \, \mathrm{J}

Hence, the kinetic energy of the object at the end of the displacement is Option C:

4 J4 \mathrm{~J}

.

Q5
An object falls from a height of 10 m10 \mathrm{~m} above the ground. After striking the ground it loses 50%50 \% of its kinetic energy. The height upto which the object can rebounce from the ground is:
A 7.5 m7.5 \mathrm{~m}
B 10 m10 \mathrm{~m}
C 2.5 m2.5 \mathrm{~m}
D 5 m5 \mathrm{~m}
Correct Answer
Option D
Solution

To solve this problem, we need to consider the principles of energy conservation and the behavior of the object during the rebound.

Let's break it down step-by-step: 1.

When the object falls from a height of

10 m10 \mathrm{~m}

, it converts its potential energy to kinetic energy at the point of impact.

The potential energy (PE) just before hitting the ground can be calculated using the formula:

PE=mghPE = mgh

where:

mm

is the mass of the object

gg

is the acceleration due to gravity (

9.8 m/s29.8 \mathrm{~m/s^2}

)

hh

is the height (

10 m10 \mathrm{~m}

) The kinetic energy (KE) of the object just before impact is equal to the potential energy it had at the height of

10 m10 \mathrm{~m}

, because of the conservation of energy principle. 2. Upon hitting the ground, the object loses

50%50\%

of its kinetic energy. Therefore, the kinetic energy after the impact is:

KEafter impact=12KEbefore impactKE_{after\ impact} = \frac{1}{2} KE_{before\ impact}

3.

The object will now rebound to a height where its kinetic energy is converted back to potential energy.

Let this height be

hreboundh_{rebound}

. The potential energy at this rebound height is:

PErebound=mghreboundPE_{rebound} = mgh_{rebound}

Since the kinetic energy after the impact is

12mgh\frac{1}{2} mgh

, we set this equal to the potential energy at the rebound height:

12mgh=mghrebound\frac{1}{2} mgh = mgh_{rebound}

By canceling out the mass

mm

and the gravitational constant

gg

, we get:

12h=hrebound\frac{1}{2} h = h_{rebound}

Substituting

h=10 mh = 10 \mathrm{~m}

, we find:

12×10 m=hrebound\frac{1}{2} \times 10 \mathrm{~m} = h_{rebound}
hrebound=5 mh_{rebound} = 5 \mathrm{~m}

Therefore, the height up to which the object can rebound from the ground is

5 m5 \mathrm{~m}

. The correct answer is: Option D

5 m5 \mathrm{~m}
Q6
At any instant of time tt, the displacement of any particle is given by 2t−12 t-1 (SI\mathrm{SI} unit) under the influence of force of 5 N5 \mathrm{~N}. The value of instantaneous power is (in SI\mathrm{SI} unit):
A 10
B 5
C 7
D 6
Correct Answer
Option A
Solution

To find the instantaneous power delivered by a force, we can use the formula:

P=F⃗⋅v⃗P = \vec{F} \cdot \vec{v}

where

PP

is the power,

F⃗\vec{F}

is the force vector, and

v⃗\vec{v}

is the velocity vector of the particle. In this question, the force exerted is given as a constant

5 N5 \, N

.

Since no direction is specified, and the displacement is given in a scalar form, we assume the force acts along the direction of displacement.

Therefore, we can treat the vectors as scalars for simplicity.

The displacement of the particle is given by:

x=2t−1x = 2t - 1

The velocity,

v⃗\vec{v}

, is the derivative of displacement with respect to time. Deriving the displacement equation with respect to time

tt

gives:

v=dxdt=ddt(2t−1)=2v = \frac{dx}{dt} = \frac{d}{dt}(2t - 1) = 2

The instantaneous power can now be calculated as:

P=F⋅v=5×2=10 WattsP = F \cdot v = 5 \times 2 = 10 \, \text{Watts}

Thus, the instantaneous power delivered by the force at any time

tt

is

10 Watts10 \, \text{Watts}

. The correct answer is Option A: 10 .

Q7
A particle moves with a velocity (5i^−3j^+6k^) ms−1(5 \hat{i}-3 \hat{j}+6 \hat{k}) ~\mathrm{ms}^{-1} horizontally under the action of constant force (10i^+10j^+20k^)N(10 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+20 \hat{\mathrm{k}}) \mathrm{N}. The instantaneous power supplied to the particle is :
A 200 W200 \mathrm{~W}
B Zero
C 100 W100 \mathrm{~W}
D 140 W140 \mathrm{~W}
Correct Answer
Option D
Solution
P=F⃗⋅V⃗P=(10i^+10j^+20k^)⋅(5i^−3j^+6k^)P=50−30+120P=140 W\begin{aligned} & P=\vec{F} \cdot \vec{V} \\ & P=(10 \hat{i}+10 \hat{j}+20 \hat{k}) \cdot(5 \hat{i}-3 \hat{j}+6 \hat{k}) \\ & P=50-30+120 \\ & P=140 \mathrm{~W} \end{aligned}
Q8
The potential energy of a long spring when stretched by 2 cm2 \mathrm{~cm} is U. If the spring is stretched by 8 cm8 \mathrm{~cm}, potential energy stored in it will be :
A 4U
B 8U
C 16U
D 2U
Correct Answer
Option C
Solution

The potential energy (U) stored in a spring when it is stretched or compressed is given by the formula:

U=12kx2U = \frac{1}{2} k x^2

where:

kk

is the spring constant,

xx

is the displacement from the spring's equilibrium position (i.e., how much the spring is stretched or compressed).

If the spring is initially stretched by

22

cm (or

0.020.02

meters, since we generally use SI units for these calculations), the potential energy stored can be represented as:

U=12k(0.02)2U = \frac{1}{2} k (0.02)^2

When the spring is stretched by

88

cm (or

0.080.08

meters), the new potential energy

U′U'

becomes:

U′=12k(0.08)2U' = \frac{1}{2} k (0.08)^2

To find the relation between

U′U'

and

UU

, we can divide the expression for

U′U'

by that of

UU

:

U′U=12k(0.08)212k(0.02)2\frac{U'}{U} = \frac{\frac{1}{2} k (0.08)^2}{\frac{1}{2} k (0.02)^2}

Simplifying this quotient gives:

U′U=(0.080.02)2=(4)2=16\frac{U'}{U} = \left(\frac{0.08}{0.02}\right)^2 = \left(4\right)^2 = 16

Thus, if the spring's displacement is increased from

22

cm to

88

cm, the potential energy stored in the spring increases by a factor of

1616

, meaning:

U′=16UU' = 16U
Q9
An electric lift with a maximum load of 2000 kg (lift + passengers) is moving up with a constant speed of 1.5 ms −-1 . The frictional force opposing the motion is 3000 N. The minimum power delivered by the motor to the lift in watts is : (g = 10 ms −-2 )
A 23000
B 20000
C 34500
D 23500
Correct Answer
Option C
Solution

F up = 2000 g + 3000 = 23000 N Minimum power P min =

F→\overrightarrow F

.

v→\overrightarrow v

P min = Fv = 23000 ×\times

32{3 \over 2}

= 34500 W

Q10
The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is
A 36 ×\times 10 7 J
B 36 ×\times 10 4 J
C 36 ×\times 10 5 J
D 1 ×\times 10 5 J
Correct Answer
Option A
Solution

Energy = Power ×\times time E = 100 ×\times 10 3 ×\times 3600 = 36 ×\times 10 7 J

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