#### MCQs on Work, Power and Energy

#### Work

**1.** Two quantities that have same SI unit are

(a) Efficiency and power (b) Efficiency and work

**(c) Work and energy** (d)
Energy and power

**Explanation:** Work and energy are not different from one another. Work is just
energy in transit. When we use some energy source to do work, we in fact,
transform one form of energy into another. Therefore, work and energy both have
same units.

**2.** In order to do work, energy is

**(a) Transferred or
converted** (b) used up (c) lost (d)
lost or transferred

**3.** SI unit for work done is

(a) Pascal **(b) Joule** (d) Newton (d)
Watt

**4.** Work in relevance
to time is related to

(a) Efficiency (b)
Force **(c) Power** (d) Momentum

**Explanation:** Power is defined as the rate of doing work, that is, the amount
of work done in one second. Therefore, the relation of work and time is power.

**5. **Work
can only be done when there is

(a) Force (b) Power **(c) Energy** (d) Efficiency

**Explanation:**
Energy possessed by a body is its ability to do work. If energy is not
available work can’t be performed.

**6.** When force is
applied on an object but object does not move, it means that

(a) No power is used **(b) no work is done** (c) work is done (d) power is used

**Explanation:** Yes, work is done when a force moves an object over a distance.
We know, Work = **F**.**d**, where **F** is the force applied and **d**
is the displacement of the object. If
the object doesn’t move at all, displacement is zero. The work done is also zero,
therefore.

**7.** When direction of applied force and
direction in which object moves are perpendicular to each other, then

(a) No power is used **(b) no work is done** (c) work is done (d) power is used

**Explanation:** As work is the scalar product of force and displacement,
therefore, it depends upon the cosine value of the angle between force and
displacement. If force and displacement are perpendicular to one another, the
angle between them is 90^{0}, whose cosine value is 0. Therefore, the
work done is also zero in such a case.

**8.** From physics point
of view, work is being done by

(a) A man carrying a heavy pile of books in stationary position (b) student solving Physics problems

(c) Student rote learning for exam **(d) none of them**

**Explanation: **Work is done only when a force acts upon somebody and causes
displacement. In all these cases, no force is exerted on somebody to cause displacement;
therefore, no work is done.

**9.** Work done by the force of friction is always

**(a) Negative** (b) zero (c) positive (d) depending on situation

**Explanation:** Friction always opposes
the motion of a body. Therefore, it is always makes an angle of 180^{0 }with
the displacement. Since cos180^{0} is -1, therefore, the work done by
friction is always negative.

**10.** When a body
slides against a rough horizontal surface, the work done by friction is

a) Positive

b) Zero

**c) Negative
**d) Constant

**Explanation:**If a force acting on a body has a component in the opposite direction of displacement, the work done is negative, when a body slides against a rough horizontal surface, its displacement is opposite to that of the force of friction. Hence work done by the friction is negative.

**11.** When a body falls
freely under gravity, then the work done by the gravity is

**a) Positive**

b) Negative

c) Zero

d) Infinity

**Explanation:** If a force acting on a
body has a component in the direction of displacement, then the work done by
the force is positive. Hence when a body falls freely under the action of
gravity the direction of the gravitational force and displacement is same. So the
work is positive.

**12.** A certain machine exerts a force of 200 newtons on a box whose
mass is 30
kilograms. The machine moves the box a distance of 20 meters along a horizontal
floor. What amount of work does the machine do on the box?

a) 120000 J (b) 300 J **(c) 4000 J** (d)
400 J

**Solution** F
= 200 N d = 20 m

W = F.d = 200*20 = 4000 J

**13.** A car travels a
distance of 15 km with a constant force of 500 N, its work done is

**(a) 7500000 J** (b) 30 J (c) 15500 J (d) 750 J

**Solution:** d = 15 km = 15000 m F = 500 N

W = 15000*500 = 7500000 J

**14.** A bus
travels with a constant force of 5000 N and work done by bus is 2500 J,
distance travelled by bus is

(a) 2 m **(b) 0.5 m** (c) 250 m (d)
12500 m

**Solution: ** Force
= 5000 N W = 2500
J

W = F d OR d = W/F

d = 2500/5000 = 0.5 m

**15.** A plane flies a
distance of 5000 m with work done of 25000 J, force applied is

(a) 20000 N (b)
10000 N (c) 125000 N **(d) 5 N**

**Solution:** d = 5000 m W = 25000 J

As F = W/d, therefore, F = 25000/5000 = 5 N

**16. **A block of mass m is moved
over a distance d. An applied force F is opposite the block displacement. How much work is done on the block by this force.

(a) mFd (b)
Fd **(c) –Fd** (d) F/d

**Explanation: **As work is the dot product of force and displacement, it depends
upon the cos value of the angle between them. When force is opposite to the
displacement, the angle between them is 180^{0}. Since cos180^{0}
= -1, therefore, the work done is negative and the correct option is (c).

**17**. When a body is lifted, the work done by
the gravitational force is.

(a)
Positive (b) Neutral (d) Depends **(d) Negative**

**Explanation:** When a body is
lifted, the work done by the gravitational force is negative. This is because
the gravitational force acts vertically downwards while the displacement is in
vertically upward direction.

**18.** For a body moving in a
circular path, the work done by the centripetal force is

(a)
Negative (b) Positive **(c) Zero** (d) Constant

**Explanation:** For a body moving in a
circular path, the centripetal force and the displacement are perpendicular to
each other. As cos 90^{0} = 0^{,} the work done by the
centripetal force is also zero.

**19.** When a coolie walks on a
horizontal platform with a load on his head, the work done by the coolie on the load is

(a) Maximum **(b) zero** (c) constant (d) mg

**Explanation:** When a coolie walks on a horizontal platform with load on his
head, he applies force in the upward direction equal to its weight. The
displacement of the load is along the horizontal direction. Thus the angle
between force and displacement is 90^{0}. Work done by the coolie on the load is zero.

**20.** A person is holding a
bucket by applying a force of 10N. He moves a horizontal distance of 5m and
then climbs up a vertical distance of 10m. Find the
total work done by him

(a) 50 J (b)
150 J (c)
500 J **(d) 100 J**

**Solution:** For horizontal distance, F
= 10N, s = 5m, θ = 90°

Work done, W_{1}=Fscosθ = 10×5×cos90° = 0.

For vertical motion, the angle between force and displacement is 0°.

So, F = 10N, s = 10m, θ=0°

Work done, W_{2}=10×10×cos0 = 100J

Total work done = W_{1}+W_{2} = 100J

Thanks sir

Your are welcome.