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

**Answer : ****Option B**

**Explaination / Solution: **

W=Fscosθ=150×20×

3√2=2598JWorkspace

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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

**Answer : ****Option D**

**Explaination / Solution: **

F⃗ =200i^+2

F⃗ =200i^+2

s⃗ =10i^+10

W=F⃗ .s⃗ =(200i^+2j^).(10i^+10j^)

W=2000+20=2020J

jj

jj

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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

**Answer : ****Option A**

**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)

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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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

**Answer : ****Option C**

**Explaination / Solution: **

angle betwwen force and displacement is

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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

**Answer : ****Option C**

**Explaination / Solution: **

W=Fscosθ=mghcos0∘W=20×10×10×1=2000J

W=Fscosθ=mghcos0∘W=20×10×10×1=2000J

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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

**Answer : ****Option C**

**Explaination / Solution: **

W=Fscosθ=mghcos0∘W=2×10−3×10×1000×1=20J

W=Fscosθ=mghcos0∘W=2×10−3×10×1000×1=20J

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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

**Answer : ****Option B**

**Explaination / Solution: **

from work kinetic energy theorm

from work kinetic energy theorm

ΔK=W12mv2−12mu2=Fs(12×6×v2)−0=12×33v2=36v2=12v=3.5m/sec

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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

**Answer : ****Option A**

**Explaination / Solution: **

ΔK=W

ΔK=W

F=12N

f=μR=μmg=0.15×6×9.8=8.82N

12mv2−12mu2=Fnets

(12×6×v2)−0=3.18×3

3v2=3.18×3

v2=3.18

v=1.8m/sec

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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

**Answer : ****Option D**

**Explaination / Solution: **

95% potential energy is converted in kinetic energy.

applying conservation of mechanical energy between horizontal and lowermost points

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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

**Answer : ****Option B**

**Explaination / Solution: **

from conservation of linear momentum initial momentum = final momentum

negative sign indicate that cars moves to the left.

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