This post has been a bit

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 = *p*^{2}/2*m* = 6^{2}/(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 = 1^{2}/(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 (*P*_{out}) of the system is *Fv = *5000×0.8 = 4000 W.

If the power input is *P*_{in} we have

efficiency *h* = *P*_{out}/* P*_{in}

Therefore, 80/100 = 4000/*P*_{in} so that *P*_{in} = 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)* **av*^{2}*ρ/*2

(b) *av*^{3}*ρ/*2

(c) *av**ρ/*2

(d) *av*^{3}*ρ*

(e)* **av*^{2}*ρ*

The mass of liquid flowing per second is *av**ρ*. Therefore, the time rate at which kinetic energy is imparted to the liquid is ½ (*av**ρ*)*v*^{2} = *av*^{3}*ρ/*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 *t* is proportional to

(a) *t*^{–2}

(b)* t*^{3}

(c) *t*^{1/2}

(d) *t*^{3/2}

(e)* t*^{–}^{3}

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, *v*^{2}/2 = (*K/m*)*t*

Therefore *v *α *t*^{1/2}.

If ‘*s*’ is the displacement, *v* = *ds/dt *so that (*ds/dt*) α *t*^{1/2}.

Integrating, *s* α *t*^{3/2} [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* = *d*^{2}*E/dv*^{2}

(e) *p* = (*dE/dv*)^{2}^{}

If the mass of the particle is *m* we have *E * = ½ *mv*^{2}

Differentiating, *dE/dv = mv = p*.

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