# Topic: Work Energy and Power (Test 1)

Topic: Work Energy and Power
Q.1
Balram exerts a steady force of magnitude 150 N on the stalled car as shown in the figure below; he pushes it a distance of 20 m. The car also has a flat tyre, so to make the car move straight. He pushes at an angle of 30 to the direction of motion. How much work does he do?

A. 2654 J
B. 2598 J
C. 2299 Jθθ=

D. 2323 J
Explaination / Solution:

W=Fscosθ=150×20×
32=2598J

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Q.2
Balram exerts a steady force of magnitude F = 200+ 2.0 on the stalled car as shown in the figure above, he pushes it a distance s = 10 + 10 m. How much work does he do?
A. 2320 J
B. 2120 J
C. 2220 J
D. 2020 J
Explaination / Solution:

F⃗ =200i^+2
s⃗ =10i^+10
W=F⃗ .s⃗ =(200i^+2j^).(10i^+10j^)
W=2000+20=2020J
jj

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Q.3
A railway porter is holding a weight of 100 kgs on his head and is standing still. What is the work done by him?
A. 0 J
B. 100 J
C. 200 J
D. 1000 J
Explaination / Solution:

W = Force x displacement in the direction of force Work done will be zero because porter is stationary (i.e. displacement is zero)

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Q.4
Balram pushes 500kg weight on a horizontal frictionless surface a distance of 10 m. The work done by gravitational force is
A. 5000 J
B. 100 J
C. 0 J
D. 200 J
Explaination / Solution:

angle betwwen force and displacement is

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Q.5
A weight of 20kg falls from a height of 10 m. The work done by the gravitational force is (Take g = 10 m )
A. 1500 J
B. 2500 J
C. 2000 J
D. 3000 J
Explaination / Solution:

W=Fscosθ=mghcos0W=20×10×10×1=2000J
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Q.6
A rain drop of mass 2.00 g falls from a height 1.00 km. It hits the ground with a speed of 50.0 m/s. What is the work done by the gravitational force? (Take g = 10 m )
A. 22 J
B. 24 J
C. 20 J
D. 26 J
Explaination / Solution:

W=Fscosθ=mghcos0W=2×103×10×1000×1=20J
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Q.7
A 6.0-kg block initially at rest is pulled to the right along a horizontal frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.
A. 5.5 m/s
B. 3.5 m/s
C. 2.5 m/s
D. 4.5 m/s
Explaination / Solution:

from work kinetic energy theorm

ΔK=W12mv212mu2=Fs(12×6×v2)0=12×33v2=36v2=12v=3.5m/sec

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Q.8
A 6.0-kg block initially at rest is pulled to the right along a horizontal, surface having a coefficient of kinetic friction of 0.15, by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.
A. 1.8 m/s
B. 0.8 m/s
C. 2.8 m/s
D. 3.8 m/s
Explaination / Solution:

ΔK=W
F=12N
f=μR=μmg=0.15×6×9.8=8.82N
12mv212mu2=Fnets

(12×6×v2)0=3.18×3
3v2=3.18×3
v2=3.18
v=1.8m/sec

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Q.9
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?
A. 5.5 m/s
B. 4.7 m/s
C. 4.9 m/s
D. 5.3 m/s
Explaination / Solution:

95% potential energy is converted in kinetic energy.

applying conservation of mechanical energy between horizontal and lowermost points

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Q.10
Consider the collision of two cars. Car 1 is at rest and Car 2 is moving at a speed of 2 m /s in the negative x-direction. Both cars each have a mass of 500 kg. The cars collide inelastically and stick together. What is the resulting velocity of the resulting mass of metal?
A. 1.2 m /s to the left
B. 1 m /s to the left.
C. 1.4 m /s to the left
D. 1.5 m /s to the left