A body moving at 2m/s can be stopped over a distance x. If its kinetic energy is doubled, how long will it go before coming to rest, if the retarding force remains unchanged

Q : A body moving at 2m/s can be stopped over a distance x. If its kinetic energy is doubled, how long will it go before coming to rest, if the retarding force remains unchanged
(A) x
(B) 2x
(C) 4x
(D) 8x

Solution : If R = resistance
Rx1 =ΔKE1 = KE . .. (i)
Rx2 = ΔKE2 =2KE . . . (ii)
Dividing (ii) by (i) 
 x2 = 2x1 = 2x

A body starts from rest with uniform acceleration and acquires a velocity V in time T. The instantaneous kinetic energy of the body the body in time t is proportional to

Q: A body starts from rest with uniform acceleration and acquires a velocity V in time T. The instantaneous kinetic energy of the body the body in time t is proportional to:
(A) (V/T)t
(B) (V2/T)t2
(C) (V2/T2)t
(D) (V2/T2)t2

Solution : KE = (1/2) mv2
= (m/2) (at)2
= (m/2) (Vt/T)2      (Since a= Δv/Δt = V/T)
Hence , K.E ∝ V2t2/T2

An electric motor creates a tension of 4500 N in hoisting cable and reels it at the rate of 2 m/s. What is the power of electric motor

Q: An electric motor creates a tension of 4500 N in hoisting cable and reels it at the rate of 2 m/s. What is the power of electric motor ?
(A) 15 KW
(B) 9 KW
(C) 225 W
(D) 9000 H.P

Solution :  P = T.v (numerically)
= 4500 × 2 = 9000 W = 9 kw

When a body of mass M slides down an inclined plane of inclination θ, having coefficient of friction μ through a distance s, the work done against friction is

Q: When a body of mass M slides down an inclined plane of inclination θ , having coefficient of friction μ through a distance s , the work done against friction is

(A)μ Mg cosθ s
(B)μ Mg sinθ s
(C)Mg (μ cosθ – sin θ)s
(D)None of the above

Solution : Work done against friction
= – work done by friction
= Mg cos θ s

An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angle theta with the vertical. Workdone by the force is

Q: An object of mass m is tied to a string of length L and a variable horizontal force is applied on it which starts at zero and gradually increases until the string makes an angle θ with the vertical. Work done by the force F is:
(A) mgL(1-sinθ )
(B) mgL
(C) mgL(1-cosθ )
(D )mgL(1+cosθ )
Sol: Work done by the external force = Δ E of the system (object)
= ΔPE    (since ΔKE=0 )
= mgh = mg (L – L cosθ )
= mgL (1 – cosθ )