Work, Energy and Power Easy Questions and Answers | Page - 6

Work, Energy and Power

A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is

1. mg(h + d) - 1 kd²
  2
1. mg(h - d) - 1 kd²
  2
1. mg(h - d) + 1 kd²
  2
1. mg(h + d) + 1 kd²
  2

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Gravitational potential energy of ball gets converted into elastic potential energy of the spring.
mg(h + d) = 1 kd²
2

Net work done = mg(h + d) - 1 kd² = 0
2

Correct Option: A

Gravitational potential energy of ball gets converted into elastic potential energy of the spring.
mg(h + d) = 1 kd²
2

Net work done = mg(h + d) - 1 kd² = 0
2

Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 m is

1. 18 J
1. 21 J
1. 26 J
1. 13 J

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Work done = area under F-d graph
= [2 × (7 - 3)] + 1 × 2 × (12 - 7)
2

= 8 + 5 = 13 J.

Correct Option: D

Work done = area under F-d graph
= [2 × (7 - 3)] + 1 × 2 × (12 - 7)
2

= 8 + 5 = 13 J.

A uniform force of (3î + ĵ) newton acts on a particle of mass 2 kg. the particle is displaced from position (2î + k̂) meter to position (4î + 3ĵ - k̂) meter. The work done by the force on the particle is

1. 6 J
1. 13 J
1. 15 J
1. 9 J

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Given : F^{^→} = 3î + ĵ
r₁^{^→} = (2î + k̂) , r₂^{^→} = (4î + 3ĵ - k̂)
r^{^→} = r₂^{^→} - r₁^{^→} = (4î + 3ĵ - k̂) - (2î + k̂)
or r^{^→} = 2î + 3ĵ - 2k̂
so work done by the given force w = F^{^→} . r^{^→}
= (3î + ĵ) . (2î + 3ĵ - 2k̂) = 6 + 3 = 9 J

Correct Option: D

Given : F^{^→} = 3î + ĵ
r₁^{^→} = (2î + k̂) , r₂^{^→} = (4î + 3ĵ - k̂)
r^{^→} = r₂^{^→} - r₁^{^→} = (4î + 3ĵ - k̂) - (2î + k̂)
or r^{^→} = 2î + 3ĵ - 2k̂
so work done by the given force w = F^{^→} . r^{^→}
= (3î + ĵ) . (2î + 3ĵ - 2k̂) = 6 + 3 = 9 J

Two similar springs P and Q have spring constants K_P and K_Q, such that K_P > K_Q. They are stretched, first by the same amount (case a,) then by the same force (case b). The work done by the springs W_P and W_Q are related as, in case (a) and case (b), respectively

1. W_P = W_Q ; W_P = W_Q
1. W_P > W_Q ; W_Q > W_P
1. W_P < W_Q ; W_Q < W_P
1. W_P = W_Q ; W_P > W_Q

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As we know work done in stretching spring w = 1 Kx²
2

where k = spring constant x = extension
Case (a) If extension (x) is same,
W = 1 Kx²
2

So,W_P > W_Q(K_P > K_Q)
Case (b) If spring force (F) is same W = F²
2K

So,W_Q > W_P

Correct Option: B

As we know work done in stretching spring w = 1 Kx²
2

where k = spring constant x = extension
Case (a) If extension (x) is same,
W = 1 Kx²
2

So,W_P > W_Q(K_P > K_Q)
Case (b) If spring force (F) is same W = F²
2K

So,W_Q > W_P

Consider a drop of rain water having mass 1 g falling from a height of 1 km. It hits the ground with a speed of 50 m/s. Take 'g' constant with a value 10 m/s². The work done by the (i) gravitational force and the (ii) resistive force of air is

1. (i) 1.25 J (ii) –8.25 J
1. (i) 100 J (ii) 8.75 J
1. (i) 10 J (ii) – 8.75 J
1. (i) – 10 J (ii) –8.25 J

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From work-energy theorem,
W_g + W_a = ∆K.E
or, mgh + W_a = 1 mv² - 0
2

= 10^–3 × 10 × 10³ + W_a = 1 × 10^–3 × (50)²
2

⇒ W_a = –8.75 J which is the work done due to air resistance
Work done due to gravity = mgh = 10^–3 × 10 × 10³ = 10 J

Correct Option: C

From work-energy theorem,
W_g + W_a = ∆K.E
or, mgh + W_a = 1 mv² - 0
2

= 10^–3 × 10 × 10³ + W_a = 1 × 10^–3 × (50)²
2

⇒ W_a = –8.75 J which is the work done due to air resistance
Work done due to gravity = mgh = 10^–3 × 10 × 10³ = 10 J