Work, Energy and Power
- A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of
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Since height is same for both balls, their velocities on reaching the ground will be same
∴ K.E1 = 1 m1v02 = m1 = 2 = 1 2 K.E2 1 m2v02 m2 4 2 2 Correct Option: D
Since height is same for both balls, their velocities on reaching the ground will be same
∴ K.E1 = 1 m1v02 = m1 = 2 = 1 2 K.E2 1 m2v02 m2 4 2 2
- 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)
- 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
- 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
- A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 m/s2)
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When the body is thrown upwards. its K.E is converted into P.E. The loss of energy due to air friction is the difference of K.E and P.E.
1 mv2 - mgh = 1 × 1 × 400 - 1 × 18 × 10 = 200 – 180 = 20 J 2 2 Correct Option: D
When the body is thrown upwards. its K.E is converted into P.E. The loss of energy due to air friction is the difference of K.E and P.E.
1 mv2 - mgh = 1 × 1 × 400 - 1 × 18 × 10 = 200 – 180 = 20 J 2 2
