Work, Energy and Power
- When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the potential energy stored in it will be
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If k be the spring constant, then U = 1 × k × (2)2 = 2 k 2 Ufinal = 1 × k × (10)2 = 50 k 2 ⇒ U = 2k = 1 Ufinal 50k 25
⇒ Ufinal = 25 UCorrect Option: A
If k be the spring constant, then U = 1 × k × (2)2 = 2 k 2 Ufinal = 1 × k × (10)2 = 50 k 2 ⇒ U = 2k = 1 Ufinal 50k 25
⇒ Ufinal = 25 U
- If the kinetic energy of a particle is increased by 300%, the momentum of the particle will increase by
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New K.E., E' = 4E
p = √2 mE and p' = √2 mE'
p' - 1 = 2 - 1 [on substrating 1 in both sides.] p p' - p × 100 = (2 - 1) × 100 = 100% p Correct Option: C
New K.E., E' = 4E
p = √2 mE and p' = √2 mE'p' - 1 = 2 - 1 [on substrating 1 in both sides.] p p' - p × 100 = (2 - 1) × 100 = 100% p
- In a simple pendulum of length l the bob is pulled aside from its equilibrium position through an angle θ and then released. The bob passes through the equilibrium position with speed
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If l is length of pendulum and θ be angular amplitude then height
h = AB - AC = l – l cos θ = l(1 – cos θ)
At extreme position, potential energy is maximum and kinetic energy is zero; At mean (equilibrium) position potential energy is zero and kinetic energy is maximum, so from principle of conservation of energy.
(KE + PE) at P = (KE + PE) at B0 + mgh = 1 mv2 + 0 2
⇒ v = √2gh = √2gl(1 – cos θ)
Correct Option: D
If l is length of pendulum and θ be angular amplitude then height
h = AB - AC = l – l cos θ = l(1 – cos θ)
At extreme position, potential energy is maximum and kinetic energy is zero; At mean (equilibrium) position potential energy is zero and kinetic energy is maximum, so from principle of conservation of energy.
(KE + PE) at P = (KE + PE) at B0 + mgh = 1 mv2 + 0 2
⇒ v = √2gh = √2gl(1 – cos θ)
- Consider a car moving along a straight horizantal road with a speed of 72 km/h. If the coefficient of static friction between road and tyres is 0.5, the shortest distance in which the car can be stopped is
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Force due to friction = kinetic energy
μ mg s = 1 mv2 2
[ Here, v = 72 km/h= 72000 = 20 m/s ] 60 × 60 or , s = v2 = 20 × 20 = 40 m 2μg 2 × 0.5 × 10
Correct Option: B
Force due to friction = kinetic energy
μ mg s = 1 mv2 2
[ Here, v = 72 km/h= 72000 = 20 m/s ] 60 × 60 or , s = v2 = 20 × 20 = 40 m 2μg 2 × 0.5 × 10
- If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is
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Initial momentum (p1) = p; Final momentum (p2) = 1.5 p and initial kinetic energy (K1) = K.
Kinetic energy (K) = p2 ∝ p2 2m or, K1 = 
p1 
2 = p 2 = 1 K2 p2 1.5 p 2.25
or, K2 = 2.25 K.
Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.
Correct Option: C
Initial momentum (p1) = p; Final momentum (p2) = 1.5 p and initial kinetic energy (K1) = K.
Kinetic energy (K) = p2 ∝ p2 2m or, K1 = p1 2 = p 2 = 1 K2 p2 1.5 p 2.25
or, K2 = 2.25 K.
Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.
