Worksheet with Solutions: Work, Power and Energy

Ans: (a)
Explanation: The SI unit of work is the Joule (symbol J). One joule is the work done when a constant force of one newton displaces an object by one metre in the direction of the force, so option (a) is correct.

Ans: (c)
Explanation: Work is a scalar quantity because it has magnitude only and no direction. In contrast, force, velocity and acceleration are vector quantities; they have both magnitude and direction. Hence option (c) is correct.

Ans: (c)
Explanation: Work is given by W = F × d × cos(θ), where θ is the angle between force and displacement. If θ = 90°, cos(90°) = 0, so W = 0. Thus the work done is zero, so option (c) is correct.

Ans: (c)
Explanation: The work-energy theorem states that the total work done by all forces acting on an object equals the change in its kinetic energy, ΔK.E. = W. Therefore option (c) is correct.

Ans: (b)
Explanation: Gravitational potential energy near Earth's surface is given by PE = mgh, where m is mass, g is the acceleration due to gravity, and h is the height above a chosen reference. Hence option (b) is correct.

Ans: displacement
Explanation: The work done by a force is calculated as the product of the force and the displacement in the direction of the force. This relationship explains how forces transfer energy to cause motion.

Ans: watt
Explanation: Power is the rate at which work is done or energy is transferred. The SI unit is the watt (W), equal to one joule per second.

Ans: kinetic
Explanation: The energy due to motion is called kinetic energy. For a body of mass m moving with speed v, K = 1/2 mv².

Ans: thermal
Explanation: When work is done against friction, mechanical energy is dissipated and converted into thermal energy (heat), raising the internal energy of the surfaces in contact.

Ans: energy
Explanation: The law of conservation of energy asserts that within an isolated system energy is neither created nor destroyed, only changed from one form to another.

Ans: True
Explanation: By definition, work requires both a force and a displacement component along the direction of that force. If there is no component of displacement along the force, no work is done.

Ans: False
Explanation: Power is a scalar quantity; it measures the rate of doing work or transferring energy and has magnitude only, not direction.

Ans: True
Explanation: When a force acts opposite to the direction of an object's displacement, it does negative work, which corresponds to a loss of kinetic energy of the object.

Ans: False
Explanation: Kinetic energy is proportional to the square of speed (K = 1/2 mv²); therefore, if speed decreases, kinetic energy also decreases.

Ans: True
Explanation: Potential energy is stored energy due to an object's position or configuration, for example gravitational potential energy due to height.

Sol: 

Ans: Work in physics is the product of a force and the displacement through which the force acts in its direction. It is given by W = F × d × cos(θ) for a force at angle θ to the displacement, and is measured in joules (J).

Ans: Energy is the capacity to do work. When work is performed on an object, energy is transferred to that object, causing motion or a change in form. Thus doing work increases the energy of the object.

Ans: The formula for kinetic energy is K = 1/2 mv², where m is the mass and v is the speed of the object. This gives the energy due to motion.

Ans: Potential energy is stored energy due to an object's position or condition. For example, a ball held at height h has gravitational potential energy equal to mgh relative to a chosen reference level.

Ans: Power is the rate at which work is done or energy is transferred. It is given by P = W / t, where W is work and t is the time taken. The SI unit is the watt (W), equal to 1 J/s.

Ans: Work, energy and power are interrelated basic concepts in mechanics:

• Work: Work is done when a force causes displacement. For a constant force, W = F × d × cos(θ), measured in joules (J). If force and displacement are parallel, W = F × d. Positive work increases an object's energy; negative work decreases it.
• Energy: Energy is the capacity to do work. Two common forms are kinetic energy (energy of motion), K = 1/2 mv², and gravitational potential energy, PE = mgh, where h is measured from a reference. Energy is conserved in isolated systems, though it may change forms.
• Power: Power is the rate of doing work or transferring energy: P = W / t. Its SI unit is the watt (W), where 1 W = 1 J/s. Higher power means more work done per unit time.
• Interrelation: The work-energy theorem states that the net work done on an object equals the change in its kinetic energy, ΔK = W. Power then tells how quickly that work is done: P = ΔK / t.
• Examples: (a) Lifting a mass m by height h against gravity requires work W = mgh, which increases the object's potential energy. (b) A car accelerating on a level road gains kinetic energy; the engine does work to increase K. The power rating of the engine determines how fast the car can gain that energy.

Ans: The work-energy theorem states that the total work done by all forces acting on a particle equals the change in its kinetic energy, ΔK = W.

• Constant forces: When a constant force F acts along the displacement d, work is W = Fd (or W = Fd cosθ for non-parallel). Example: pushing a box on a frictionless surface with constant force increases its kinetic energy by W.
• Variable forces: For a force that varies with position, work is found by integration: W = ∫ F(x) dx between limits. Example: stretching a spring (Hooke's law, F = -kx), the work done in stretching from x = 0 to x = x₁ is W = ∫0 (-kx) dx = -(1/2) k x₁², which equals the negative change in the spring's potential energy. The work-energy theorem still applies: the net work equals the change in kinetic energy regardless of whether forces are constant or variable.
• Application: For a falling object, gravity does positive work, converting potential energy into kinetic energy; air resistance does negative work, converting mechanical energy into heat. The theorem allows calculation of final speeds without solving the full equation of motion in many cases.

Ans: (a)

Explanation:
(i) Assertion: Work done by a conservative force is path independent.
(ii) Reason: Work done by a conservative force in a closed loop is zero.
(iii) Justification: For conservative forces (for example, gravity or ideal spring force), the work between two points depends only on the initial and final positions. If a path is closed, initial and final positions coincide, so the net work is zero. Therefore R is true and it correctly explains A.

Ans: (a)

Explanation:
(i) Assertion: A spring force is a conservative force.
(ii) Reason: The work done by a spring force depends only on the displacement, not on the path taken.
(iii) Justification: Hooke's law gives F = -kx for an ideal spring. The work done in moving a mass attached to a spring from x₁ to x₂ is ∫_x1^x2 (-kx) dx = -(1/2)k(x₂² - x₁²), which depends only on the end positions. Thus the spring force is conservative and R correctly explains A.

Ans: (d)

Explanation:
(i) Assertion: In an inelastic collision, total kinetic energy is conserved. - This is false; in inelastic collisions some kinetic energy is converted into other forms (heat, deformation).
(ii) Reason: Total linear momentum is conserved in all types of collisions. - This is true provided no external forces act on the system.
(iii) Justification: Momentum conservation holds for isolated systems, but kinetic energy is conserved only in elastic collisions, not in inelastic ones. Therefore A is false and R is true.

Ans: Final K.E. ≈ 0.50 J (approximately)

Sol:
Given: mass m = 1 kg, initial speed v_i = 2 m/s.
Initial kinetic energy K_i = 1/2 × m × v_i² = 1/2 × 1 × 2² = 2 J.
The retarding force is F(x) = -0.5 / x (N) for x from 0.1 m to 2.01 m.
Work done by the force, W = ∫_0.1^2.01 F(x) dx = ∫_0.1^2.01 (-0.5 / x) dx = -0.5 × ln(x) |_0.1^2.01 = -0.5 × ln(2.01 / 0.1).
Compute ln(2.01 / 0.1) = ln(20.1) ≈ 3.0007, so W ≈ -0.5 × 3.0007 ≈ -1.50035 J.
By the work-energy theorem, K_f = K_i + W = 2 + (-1.50035) ≈ 0.49965 J ≈ 0.50 J (to two significant figures). Final kinetic energy ≈ 0.50 J.

Ans: Final speed ≈ 63.2 m/s

Sol:
Mass m = 50 g = 0.05 kg. Initial speed u = 200 m/s.
Initial kinetic energy K_i = 1/2 × m × u² = 1/2 × 0.05 × 200² = 0.025 × 40 000 = 1000 J.
Bullet retains 10% of initial K.E., so final kinetic energy K_f = 0.10 × 1000 = 100 J.
Use K_f = 1/2 × m × v² ⇒ v² = 2K_f / m = 2 × 100 / 0.05 = 4000.
Thus v = √4000 ≈ 63.2456 m/s ≈ 63.2 m/s (approximately).