1st PUC Physics Question Bank Chapter 6 Work, Energy and Power

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Karnataka 1st PUC Physics Question Bank Chapter 6 Work, Energy and Power

The Scalar Product

Question 1.
Define scalar product of two vectors. Give an example.
Answer:
The scalar product or dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as
\(\vec{A} \cdot \vec{B}\) = AB cos θ
where θ is the angle vectors between vectors \(\vec{A}\) and \(\vec{B}\)
Example:
work = \(\vec{F} \cdot \vec{S}\) = FS cos θ
The dot product of two vectors result in a scalar quantity.
Note:
(i) \(\vec{A} \cdot \vec{B}\) = \(\vec{B} \cdot \vec{A}\) i.e., scalar product is commutative.
(ii) \(\vec{A} \cdot(\vec{B}+\vec{C})=\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{C}\) i.e., scalar product is distributive.
(iii) If θ = 0° then \(\vec{A} \cdot \vec{B}\) = AB and is maximum.
(iv) If θ = 90° then \(\vec{A} \cdot \vec{B}\) = 0 and is minimum.

Question 2.
Define workdone. Mention its SI unit.
Answer:
Work is said to be done when force is applied on the body and the body displaces in the direction of applied
force. W = \(\vec{F} \cdot \vec{S}\)
S.I. unit of work is joule (J)

Question 3.
Define 1 joule.
Answer:
w.k.t. W = FS cos θ , if F = 1 N, S = 1 m and θ = 0° then W = 1 J.
Work is said to be 1 J when a force of 1 N displaces the object through 1 m in the direction of force.

Question 4.
Define gravitational unit of work.
Answer:
Gravitational unit of work is kilogram meter.
Work is said to be 1 kgm when a force of 1 kg wt. displaces a body through 1 m in its direction.
1 kg m = 1 kg wt = 9.81 J

Work Done by a Constant Force

Question 1.
Mention the expression for work done by a constant force. When the work is said to be (a) positive (b) negative (c) zero.
Answer:

Consider a force F is acting on a body at an angle θ in the direction of displacement.
Now is S is the displacement of the body, then
W = FS cos θ
W = \(\vec{F} \cdot \vec{S}\)
(a) If θ is < 90° then work done is positive
(b) If θ > 90° then work is done against the force, and the work done is negative.
(c) If θ = 90° Now work done by the force is zero.
Example: Work done in moving a particle in a circle with constant speed is zero.

Question 2.
What is the work done by a weightlifter holding 150 kg of mass on his shoulder for 30 s?
Answer:
Zero

Work Done by a Variable Force

Question 1.
Obtain an expression for work done by a variable force,
Answer:
Consider a variable force F (x) is acting on a body and it is displaced from A(x₁) to B(x₂) in the direction of force.

The total work done can be calculated as follows.
Let dW be the work done in displacing the body by small distance dx. Now dW = F(x)dx
Total work done is obtained by integrating between limits x₁to x₂.
W = \(\int d w=\int_{x_{1}}^{x_{2}} F(x) d x\) = Area under curve between x₁to x₂.

Question 2.
What does the area under force and displacement represents?
Answer:
The area under force and displacement represents the work done.

Energy

Question 1.
Define energy and mention the types of mechanical energy.
Answer:
Energy is the capacity of doing work.
Its SI unit is joule (J)
There are two types of mechanical energy
(a) Potential energy (b) Kinetic energy

Question 2.
What is kinetic energy?
Answer:
It is the energy possessed by a body by virtue of its motion.

Question 3.
Derive an expression for kinetic energy.
Answer:
Let a body of mass m is initially at rest is acted upon by a constant force F.

Let it acquires final velocity v after displacement ‘S’
then v² – u² =2as
v² – 0 = 2as
a = acceleration
a =\(\frac{v^{2}}{2 s}\) …. (1)
but F=ma
F = m\(\frac{v^{2}}{2 s}\)
and W=FS
W = \(\frac{m v^{2}}{2 S}(S)=\frac{1}{2} m v^{2}\)
This work done appears as kinetic energy
∴ KE = \(\frac{1}{2}\) mv²

Question 4.
Obtain the relation between kinetic energy and linear momentum.
Answer:
We know that
KE = \(\frac{1}{2}\) mv²
dividing and multiply (1) by m
k.E =\(\frac{1}{2} m v^{2} \frac{m}{m}\)
k.E = \(\frac{m^{2} v^{2}}{2 m}\)
k.E = \(\frac{p^{2}}{2 m}\) [∵ P = mv]
This work done appears as kinetic energy

Question 5.
State and prove work energy theorem for a constant force.
Answer:
Statement: The work done by net force in displacing a body is equal to change in its kinetic energy.
Let a body of mass m is moving with initial velocity u is applied with a force F so that its final velocity is v in displacing the body by ‘S’.
Now v² – u² = 2as
where a = acceleration

∴work =change in kinetic energy

Question 6.
Define potential energy and mention the types of potential energy.
Answer:
The energy possessed by a body by virtue of its position is known as potential energy.
Mechanical potential energies are of two types :

1. Gravitational potential energy
2. Elastic potential energy

Question 7.
What is gravitational potential energy? Obtain an expression for it.
Answer:
The energy possessed by a body (mass) due to its position in the gravitational field is called as gravitational potential energy.
Let a force F is required to raise a body of mass ?fl to height h against Gravitational
force, then F
F=mg .. (1)
The work done by the force.
W= \(\vec{F} \cdot \vec{h}\)
W = Fhcosθ But θ = 0°, cos 0° = 1
W = mgh
This work done is stored as potential energy U = mgh

Question 8.
Mention the relation between gravitational force and gravitational potential energy.
Answer:
If F is the gravitational force and gravitational potential energy is a function of height U (h), then F = \(\frac{-d}{d h} U(h)\) , negative sign indicates that F is opposite to displacement h.

Question 9.
What is elastic potential energy? Obtain an expression fornotential energy of a spring.
Answer:
The energy possessed by a body due to its configuration is called elastic potential energy.
Let a spring of spring constant ‘k’ is elongated or compressed by distance ‘x’.
Now the spring develops a force called spring force F.
This force F is always opposite to displacement ‘x’.
Restoring force F = -kx …….(1)
Now external force is equal and opposite to restoring force
F_ext =-F
∴ F_ext = kx
If dW is small work done in small displacement ‘dx’

dW = F_extdx = kx dx
∴ Total work in displacing spring through ‘x’ is

This work done will be stored as potential energy spring
U = \(\frac{1}{2}\) kx²

Conservative, Non-conservative Forces

Question 1.
What do you mean by conservative forces? Mention its properties.
Answer:
A force is said to be conservative if work done by or against the force in moving a body depends only on initial and final positions of the body.
Properties
(a) Work done depends on initial and final positions
(b) Work done does not depends on nature of the path followed by the body
(c) Work done in moving a body through a round trip is zero.
Example .
Gravitational force
Electrostatic force

Question 2.
State and prove the law of conservation of mechanical energy for a body falling freely under gravity.
Answer:
Statement: The total mechanical energy of the system is conserved, is work is done by or against the conservative force.
Consider a body is falling freely from the height ‘h’.
At position ‘A’
KE = 0
PE = mgh
∴ TE = KE + PE = mhh ….. (1)
At position ‘B’
PE = mg(h-x)
KE = \(\frac{1}{2}\)mv²
v² = u² + 2 gx (u = 0)
v² = 2 gx
K.E. = \(\frac{1}{2}\)m(2gx) -mgx
TE = KE + PE
TE = mgx + mgh – mgx
TE = mgh ………….. (2)
At position C
PE = 0
KE= \(\frac{1}{2}\)mv²
v² = u² = 2gh (u-0)
v = 2 gh
∴ K.E. = \(\frac{1}{2}\)m(2gh) = mgh
TE = 0 + mgh = mgh …… (3)
From (1), (2) and (3) we observe that total mechanical energy is conserved at every point.

Question 3.
Mention various forms of energy
Answer:
(a) Heat energy
(b) Chemical energy
(c) Electrical energy
(d) Nuclear energy

Question 4.
State and explain mass – energy equivalence.
Answer:
According to Einstein mass can be converted into energy and energy can be converted into mass E = mc². Where m is the mass c is the speed of light E = me2 is called mass energy equivalence.

Question 5.
State the principle of conservation of energy.
Answer:
According to this principle, energy can neither be created nor, destroyed, but it can be transformed from one form to another. Total energy is always conserved in a system.

Question 6.
Define power and mention its SI unit and practical unit.
Answer:
Power is the time rate of doing work.
P = \(\frac{\text { work }}{\text { time }}\)
SI unit of power is watt (W)
Practical units of power are
(a) 1 kilowatt (1 kW) = 1000 W
(b) 1 Horse power = 746 W
(c) 1 mega watt (1 MW) = 10⁶ W

Question 7.
What are non-conservative forces? Give example.
Answer:
A force is said to be non-conservative if work done by the force or against the force depends upon the path followed by a body.
Example: frictional force, viscous force.

Collisions

Question 1.
What do you mean by collisions? Mention the types.
Answer:
Collision is the interaction between two bodies for a short interval of time. Because of collision there will be change in momentum and energy.
There are two types:

1. Elastic collision
2. In elastic collision

Question 2.
What is elastic collision? Give example.
Answer:
A collision is said to be elastic if there is no loss of kinetic energy. OR It is the collision in which both momentum and kinetic energy are conserved.
Example: Collision between atoms in a gas.

Question 3.
What is in inelastic collision? Give example.
Answer:
It is the collision in which only momentum is conserved and there is loss of kinetic energy.
Example: Collision between two sphere made of soft clay, collision between soft clay and wall, if a bullet is fired at a block of wood and get embedded into it.

Question 4.
Derive expressions for final velocities for particles (bodies) under going elastic collisions.
Answer:
Consider two bodies having masses m₁ and m₂ collide elastically and move in same direction.
We know that momentum is conserved.

Question 5.
Discuss the elastic collision for two bodies in two dimensions.
Answer:
Consider two bodies having masses m₁ m₂ collide each other and move in different directions by making angles θ₁ and θ₂ with x-axis.

Let u₂ = 0
According to conservation of momentum Along x-axis
m₁u₁ =m₁v₁ cos θ₁ + m₂vcos θ₂ Along y-axis
0 = m₁v₁ sin θ₁ + m₂vsin θ₂
For elastic collision. KE is also conserved.
\(\frac{1}{2}\)m₁u₁² = \(\frac{1}{2}\)m₁v₁² + \(\frac{1}{2}\) m₂v₂²

Question 6.
Derive an expression for loss in kinetic energy for two bodies colliding in-elastically.
Answer:
Consider two bodies having masses m₁ and m₂ collide in elastically and move together.

We know that momentum is conserved.
∴ m₁u₁ + 0 = ( m₁ +m₂ )v
v = \(\left(\frac{m_{1}}{m_{1}+m_{2}}\right) u_{1}\) …. (1)
But KE is not conserved.
Therefore loss in KE is

Question 7.
Define co-efficient of restitution and mention expression for it.
Answer:
It is a measure of the degree of elasticity of a collision and it is defined as the ratio of relative velocity after collision to the relative velocity before collision.
e = \(\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\)
e = 1 for perfect elastic collision.

Numerical Problems

Question 1.
A cyclist comes to a skidding stop in 10m. During this process, the force on the cycle due to the road is 200 N and is directly opposed to the motion, (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?
Answer:
(a) Work done by the road on the cycle is work done by the stopping frictional force on the cycle.
W₁ = F s cos θ = (200) (10) cos 180° = – 200 J
This is obviously negative

(b) From Newton’s third law, an equal and opposite force acts on the road due to the cycle.
∴ Force on the road, F = 200 N.
But the road undergoes no displacement. Therefore, work done by cycle on the road is zero.

Question 2.
A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which trunk has been moved is 20 m. Plot the force applied by the woman and the frictional force, which is 50 N against the distance. Calculate the work done by the two forces over 20 m.
Answer:
The graph of force applied by woman and the frictional force against the distance is shown in Fig.
Work done by the force applied by woman
W₁ = area of rectangular OABD + area of trapezium BCED
= OA × OD + \(\left(\frac{B D+C E}{2}\right)D E\)
= 100 × 10 + \(\left(\frac{100+50}{2}\right) 10\) = 1000 + 750 = 1750 J
Work done by frictional force
W₂ = – area by frictional force = – OG × OE = -50 × 20 = – 1000 J

Question 3.
A man moves on a straight horizontal road with a block of mass 2 kg in his hand. If he moves a distance of 40 m with an acceleration of 0.5 ms^-2. Calculate the work done by the man on the block during motion.
Answer:
m = 2 kg, S = 40m, a = 0.5 ms^-2, θ = 0°
W = ES cos θ = FS cos 0° = FS
Since, F = ma W = maS
= 2 × 0.5 × 40 = 40 J

Question 4.
A 60 kg man runs up a flight of stairs 3 m high in 2 s. Find his average power.
Answer:
Work done = mgh
= 60 × 9.8 × 3
= 1764 J
Power = \(\frac{\text { work done }}{\text { time }}\)
= \(\frac{w}{t}=\frac{1764}{2}\)
= 882 W.

Question 5.
A body of mass 2 kg is projected at 20 ms^-1 at an angle of 60° above the horizontal. Calculate the power due to gravitational force at the highest point.
Anst P = FV cosθ
P = FV cos 90°
P = 0

Question 6.
A force \(\vec{F}\) = (2.0î – 6.0ĵ)N is applied on a body which is sliding over a floor. If the body is displaced through (-3j) m, how much work is done by the force.
Solution:
\(\vec{F}\) = (2.0î – 6.0ĵ)N, \(\vec{S}\) = -3î
W = \(\vec{F} \cdot \vec{S}\) = (2.0î – 6.oĵ}) (-3î)
= -6 + 0 = -6J

Question 7.
A cyclist comes to a skidding stop in 10m. During this process, the force on the cycle due to the road is 200 N and is directly opposite to the motion, (a) How much work does the road do on the cycle? (b) How much work does the cycle do on the road?
Answer:
S = 10 m,F = -200 N
(a) W = FS = – 200 × 10 = – 2000 J.
Thus road does 2000 J of work on the cycle.

(b) According to Newton’s 3rd law of motion, to every action there is equal and opposite reaction. Therefore force acted on the road is 200 N.
However, the road undergoes no displacement. Thus, work done by cycle on the road is zero.

Question 8.
A particle moves along the x-axis from x = 0 to x = 5m under the influence of a force given by f = 7 – 2x + 3x². Calculate the work done.
Answer:
F = 7 – 2x + 3xsup>2

Question 9.
Calculate the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to a vertical height of 10cm. Given g = 9.8ms^-2.
Answer:
h = 10 cm = 0.1m g = 9.8 ms^-2
P.E. = mgh = m × 9.8 × 0.1 = 0.98 mJ
Since at mean position, whole of the P.E. is converted into kinetic energy
∴ K.E. = P.E.
∴ \(\frac{1}{2}\) mv² =0.98 mJ or v² = \(\sqrt{2 \times 0.98}\) =1.4 ms^-1

Question 10.
A ball is dropped vertically from rest at a height of 12m. After striking the ground it bounces back to a height of 9 m. What percentage of K.E. does it lose on striking the ground?
Answer:
h₁ = 12 m, h₂ = 9 m
Let m be the mass of ball.
Initial energy = mgh₁ = 12 mg
Final energy = mgh₂ = 9 mg
Loss in K.E. = mgh₁ – mgh₂
= 12 mg – 9 mg = 3 mg
∴ Percentage loss in K.E.
= \(\frac{\text { Loss in K.E. }}{\text { Initial energy }}\) x 100
= \(\frac{3 \mathrm{mg}}{12 \mathrm{mg}}\) x 100 = 25%

Question 11.
Calculate the decrease in mass of 1 g of water at 0°C, when it turns into ice at 0°C. Given latent heat of ice = 80 cal g^-1.
Answer:
m= 1g, L = 80 cal g^-1
E = mL
= 1 × 80 = 80 cal
∴ E = 80 × 4.2 = 336 J If Am is the decrease in mass, then
E = ∆mc²
336 = ∆m(3 × 10⁸)²
∴ ∆m = \(\frac{336}{\left(3 \times 10^{8}\right)^{2}}\) = 3.74 × 10^-15 Kg

Question 12.
In a ballistics demonstration, a police officer fires a bullet of mass 50.0 g with speed 200 ms^-1 on soft plywood of thickness 2-00 cm. The bullet emerges only with 10% of its initial kinetic energy. What is the emergent speed of the bullet?
Answer:

Note that K.E. is reduced by 90% but speed is reduced by nearly 68%.

Question 13.
A block of mass m = 1 kg moving on a horizontal surface with speed v_i = 2 ms^-1 enters a rough patch ranging from x = 0-10 to m to x – 2 01 m. The retarding force Fr on the block in this range is inversely proportional to x over this range F_r= \(-\frac{k}{x}\) for 0 1 < x < 2 01 m = 0 for x < 0 1m and x > 2.01 m
where k= 0-5 J. What is the final K.E and speed v_f of the block as it crosses the patch? [NCERT Solved Example]
Answer:
Here, m = 1 kg, v_i = 2 ms^-1; k = 0-5 J
Initial K.E., Kt = \(\frac{1}{2}\) mv_i² = \(\frac{1}{2}\) × 1(2)² = 2 J
Work done against friction

W= – 0.5 × 2.303 log₁₀ 20.10 = – 0.5 × 2.303 × 1.303 =-1.5 J
∴ Final K.E., K_f=K_i + W = 2.0 – 1.5 J = 0.5 J

Question 14.
To stimulate car accidents, the auto manufacturers study the collisions of moving cars with mounted springs of different spring constants. Consider a typical simulation with a car of mass 1000 kg moving with a speed of 18.0 km on a smooth road and colliding with a horizontally mounted spring of spring constant 6-25 x 103Nm^-1. What is the maximum compression of the spring? [NCERT Solved Example]
Solution:
Here, m = 1000 kg, v = 18.0 km/h = 5 ms^-1, k = 6.25 × 10³ Nm^-1, x_m = ?
At maximum compression of the spring, the K.E. of the car is converted entirely into P.E. of the spring.

Question 15.
Consider example 15. Taking the co of friction, p to be 0.5, calculate the maximum compression of the spring.
Solution:
In the presence of friction, both the spring force and the frictional force / act so as to oppose the compression of the spring.
Work done by the two forces
w = \(-\frac{1}{2}\)kx²_m – f(x_m)
w = \(-\frac{1}{2}\)kx²_m – μmg(x_m) …. (∵ f = μR = μ mg
and change in K.E ∆K =K_f – KL_i = 0 – \(\frac{1}{2}\) mv²
Applying work energy theorem, ∆K = W

Putting the numerical values and solving we get x_m = 1.35 m

Question 16.
A light body and a heavy body have same K.E. Which is having more momentum and why?
Answer:
Let m₁ and m₂ be the masses of lighter and heavier body and s₁ and v₂ be their respective velocities.


Momentum of lighter body, p₁= m₁v₁
And momentum of heavier body , p_h = m_hv_h

Question 17.
The human heart discharges 75 ml of blood at each beat against a pressure of 0.1 m of Hg. Calculate the power of the heart assuming that the pulse frequency is 80 beats per minute. Given, density of mercury = 13.6 x 10³ kg/m³.
Answer:
Here, Volume of blood discharged / beat,
V = 75ml = 75 × 10^-6m³
Pressure, P = 0.1 m of Hg
P = 0.1 × (13.6 × 10³) × 9.8 Nm^-2 [from P = hp g]
Work done / beat = PV
Total work done in 80 beats = PV × 80

Question 18.
An elevator which can carry a maximum load of 1800 kg (elevator + passengers) is moving up with a constant speed of 2 ms^-1. The frictional force opposing the motion is 4000 N. Determine the minimum power delivered by the motor to the elevator in watt and in horse power.
Answer:
Here, m = 1800 kg
Frictional force,/= 4000 N ; uniform speed, v = 2.0 ms^-1
Downward force on elevator is
F = mg+ f -1800 × 10 + 4000 = 22000 N
The motor must supply enough power to balance this force. As P = F × v
∴ P = 22000 × 2 = 44000 watt
P = \(\frac{44000}{746}\) h.p = 59H.P

Question 19.
A standard car develops 40 H.P. Find the maximum speed the car can attain against a resistance of 20 kg wt. due to air and friction. Given efficiency of the engine is 25%. 1 H.P. = 746 W and g = 10 ms^-2.
Answer:
Here, P = 40 H.P. = 40 × 746 watt
v = ? F = 20 kg wt. = 20 × 10 N
η = 25%, g = 10 ms^-2
As η = \(\frac{\text { output power }}{\text { input power }}\)
∴ output power = input power x η = 40 x 746 x \(\frac{25}{100}\) watt = 7460 watt
As output power = force x velocity
∴ velocity = \(\frac{\text { output power }}{\text { force }}\)
v = \(\frac{7460}{20 \times 10}\) = 37.3ms^-1

Question 20.
A plastic ball is dropped from a height of 1 m and rebounds several times from the floor. If 1.3 elapses from the moment it is dropped to the second impact with the floor, what is the co-efficient of restitution?
Answer:
Here, h = 1 m, u = 0.
Velocity with which the ball strikes the floor for the first time, v₁ = \(\sqrt{2 g h}=\sqrt{2 \times 9 \cdot 8 \times 1}\) = 4.427 ms^-1
Time taken by the ball to strike the floor for the first time, t₁= \(\frac{v_{1}}{g}=\frac{4 \cdot 427}{9 \cdot 8}\) = 0.452 s
Time taken by the ball between first and second impact with the floor is
t₂ = 1.3 – t₁ = 1.3 – 0.452 = 0.848 s
Time taken by the ball for upward journey after first impact = \(\frac{t_{2}}{2}=\frac{0.848}{2}\) = 0.424 s
upward velocity after first impact v₂ = g x \(\frac{t_{2}}{2}\) = 9.8 × 0.424 = 4.155 ms^-1
Co-efficient of restitution, e = \(\frac{v_{2}}{v_{1}}=\frac{4 \cdot 155}{4 \cdot 427}\) = 0.939

Question 21.
A ball of 0.1 kg makes an elastic head on collision with a ball of unknown mass that is initially at rest. If the 0.1 kg ball rebounds at one third of its original speed, what is the mass of the other ball?
Answer:
Here, m₁ = 0.1 kg ; m₂ = ?
u₂ = 0 Let u₁ = u ;
.’. v₁ = \(-\frac{u}{3}\)

Question 22.
A and B are two particles having the same mass rn.Ais moving along X-axis with a speed of 10 ms^-1 and B is at rest. After undergoing a perfectly elastic collision with B, particle A gets scattered through an angle of 30°. What is direction of motion of B, and the speeds of A and B, after the collision?
Answer:
Fig. shows the particles A and B before collision ; and the two particles after perfectly elastic collision in two dimensions.

Here, u = 10 ms^-1 θ = 30°,Φ = ?
v₁ = ? v₂ = ?
As is known from Art. 4.24,
special case θ + Φ = 9O°
∴ Φ = 90° – 0 = 90° – 30° = 60°
Using law of conservation of Unear momentum along X-axis.
u = v₁ cosθ + v₂ cosθ
10 = v₂ cos 30° + v₂cos 60° – \(\frac{v_{1} \sqrt{3}}{2}+\frac{v_{2}}{2}\)
or 20 = √3 v₁ + v₂ …. (1)
Again, using law of conservation of Unear momentum along Y-axis.

From (i), 20 = √3v₂√3 + v2 = 4v₂
v_s = \(\frac{20}{4}\) = 5 ms^-1
From (ii), v₁ = √3v₂ = √3 × 5 = 1.732 × 5msv =8 66 ms^-1