Life is like riding a bicycle.  To keep your balance you must keep moving.”
Albert Einstein

Sunday, November 30, 2008

AP Physics B & C- Additional Practice Questions (MCQ) on Work, Energy & Power

This post has been a bit delayed because of a ‘boot.com’ virus infection in my system. Fortunately I could disinfect the system without much delay.

A few multiple choice questions on work energy and power were discussed in the last post. Here are a few more multiple choice questions in this section:

(1) A particle of mass 2 kg moving along the positive x-direction has a constant momentum of 6 kg ms–1. If a constant force of 2 N is applied on the particle for 1 s along the negative x-direction, the change in the kinetic energy of the particle is

(a) decrease of 4 J

(b) increase of 4 J

(c) decrease of 8 J

(d) increase of 4 J

(e) decrease of 9 J

The initial kinetic energy of the particle = p2/2m = 62/(2×2) = 9 J.

The force F of 2 N actng on the particle for the time t equal to 2 s imparts a momentum Ft equal to 4 kg ms–1 along the negative direction. Since the initial momentum of the particle is along the positive x-direction, the resultant momentum is 6 – 4 = 2 kg ms–1.

Therefore, the final kinetic energy of the particle = 12/(2×2) = 1 J.

The change in the kinetic energy of the particle is 1 – 9 = – 8 J

The kinetic energy of the particle is thus decreased by 8 J.

(2) An elevator motor creates a tension of 5000 N in a hoisting cable and reels it at 0.8 ms–1. If the efficiency of the elevator system is 80%, the power input to the motor is

(a) 2 kW

(b) 4 kW

(c) 6 kW

(d) 8kW

(e) 10 kW

The power output (Pout) of the system is Fv = 5000×0.8 = 4000 W.

If the power input is Pin we have

efficiency h = Pout/ Pin

Therefore, 80/100 = 4000/Pin so that Pin = 5000 W = 5 kW.

(3) A liquid of density ρ is being continuously pumped through a pipe of area of cross section a. If the speed of the liquid through the pipe is v, the time rate at which kinetic energy is being imparted to the liquid is

(a) av2ρ/2

(b) av3ρ/2

(c) avρ/2

(d) av3ρ

(e) av2ρ

The mass of liquid flowing per second is avρ. Therefore, the time rate at which kinetic energy is imparted to the liquid is ½ (avρ)v2 = av3ρ/2

The above questions will be beneficial for AP Physics B as well as C. The following questions are for AP Physics C aspirants:

(4) A machine delivering constant power moves an object along a straight line. The displacement of the object in time t is proportional to

(a) t–2

(b) t3

(c) t1/2

(d) t3/2

(e) t3

Since the power is the product of force and velocity we have

Fv = K where K is a constant.

If m is the mas and v is the velocity of the object, the above equation can be written as

M(dv/dt)v = K so that vdv = (K/m)dt

Intrgrating, v2/2 = (K/m)t

Therefore v α t1/2.

If ‘s’ is the displacement, v = ds/dt so that (ds/dt) α t1/2.

Integrating, s α t3/2 [Option (d)].

(5) The velocity v, momentum p and the kinetic energy E of a particle are related as

(a) p = dv/dE

(b) p = dE/dv

(c) p = (dE/dv)1/2

(d) p = d2E/dv2

(e) p = (dE/dv)2

If the mass of the particle is m we have E = ½ mv2

Differentiating, dE/dv = mv = p.

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