C LEVEL PROBLEMS

1)      A 900-N crate rests on the floor. How much work is required to move it at a constant speed (a) 6.0-m along the floor against a frictional force of 180 N, and (b) 6.0-m vertically?

SOLUTION:

      (a)  Because there is no acceleration, the horizontal applied force must have the same magnitude as the friction force. And, since this force is applied in the direction of the displacement, the angle between the force and the direction of displacement, q = 0. Thus, 

                  W = F s cos q = (180 N)(6.0 m)(1)  =1.1 ´ 103 J.

     

      (b)  Because there is no acceleration, the vertical applied force must have the same magnitude as the weight. And, since this force is applied in the direction of the displacement, the angle between the force and the direction of displacement, q = 0. Thus,

                  W = F s cos q = mg s cos q = (900 N)(6.0 m)(1) =5.4 ´ 103 J.

2)      At room temperature, an oxygen molecule, with a mass of 5.31 x 10-26 kg, typically has a kinetic energy of about 6.21 x 10-21 J. How fast is it moving?

SOLUTION:

We find the speed from 

            k = ½ mv2;

            6.21 ´ 10–21 J = ½ (5.31 ´ 10–26 kg) v2, which gives v = 484 m/s.

3)      How much work is required to stop an electron (m = 9.11 x 10-31 kg) which is moving with a speed of 1.90 x 106 m/s?

SOLUTION:

The work done on the electron decreases its kinetic energy since it stops the electron.  Therefore, we can use the work-kinetic energy theorem:

            W = Dk = ½mv2 – ½mv02 = 0 – ½ (9.11 ´ 10–31 kg)(1.90 ´ 106 m/s)2 = – 1.64 ´ 10–18 J.

4)      A spring has a spring constant, k, of 440 N/m. How much must this spring be stretched in order to store 25 J of potential energy?

SOLUTION:

The potential energy of the spring is zero when the spring is not stretched (x= 0). For the stored potential energy, we have

            pe = ½kxf2– 0;

            25 J = ½ (440 N/m) xf2 – 0, which gives xf= 0.34 m.

5)      By how much does the potential energy of a 64-kg pole-vaulter change if his center of mass rises about 4.0-m during the jump?

SOLUTION:

For the potential energy change we have

            DU = mg Dy = (64 kg)(9.80 m/s2)(4.0 m) =2.5 ´ 103 J.

6)      A novice skier, starting from rest, slides down a frictionless 35.00 incline whose vertical height is 125 m. How fast is she going when she reaches the bottom?

SOLUTION:

We choose the potential energy to be zero at the bottom (y = 0). Because there is no friction and the normal force does no work, energy is conserved, so we have

            Ei = Ef

            ki + Ui = kf + Uf ;

            ½mvi2 + mgyi = ½mvf2 + mgyf ;

            ½m(0)2 + m(9.80 m/s2)(125 m) = ½mvf2 + m(9.80 m/s2)(0), which gives vf = 49.5 m/s.

      This is 180 km/h! It is a good thing there is friction on the ski slopes.

7)      How long will it take a 1750-W motor to lift a 285-kg piano to a sixth-story window 16.0-m above?

SOLUTION:

The amount of work required is the increase in potential energy: W = mg Dy.

      We find the time from

            P = W/t = mg Dy/t;

            1750 W = (285 kg)(9.80 m/s2)(16.0 m)/t, which gives t = 25.5 s.

8)      If a car generates 18 hp (1 horse power = 746 Watts) when traveling a steady 90 km/h, what must be the average force exerted on the car due to friction and air resistance?

SOLUTION:

We find the equivalent force exerted by the engine from

            P = Fv;

            (18 hp)(746 W/hp) = F(90 km/h)/(3.6 ks/h), which gives F = 5.4 ´ 102 N.

      At constant speed, this force is balanced by the average retarding force, which must be5.4 ´ 102 N.