PHYSICS 100A #85014

FALL 2003 QUIZ 5

  1. A 30-N force exerted on a 20-kg cart with an angle of 30° to the horizontal as shown. After the cart was pushed 50m horizontally from rest, it gained a speed of 5m/s. What was the coefficient of kinetic friction between the cart and the ground? Use work-energy principle to solve the problem.
    Solution. From the force diagram we find the friction force:

                   y-component: FN - Fsinq - mg = 0;

          We have Ffr = mkFN = mk(Fsinq + mg).

          From work-energy principle he have

                  WNC = DKE = 1/2mvf2  - 1/2mvf2;

                 (Fcosq - Ffr)d = 1/2mvf2  - 1/2mvf2;

                 {(30N)cos30° - mk[(30N)sin30° + (20kg)(9.80m/s2)](50m)

                  = 1/2(20kg)(5m/s)2 - 0,

          which gives mk = 0.0994.

    		  
  2. A spring of spring constant 900N/m is compressed 25cm from its unstretched position. What is the change of the potential energy of the system?

    Solution. We find the change of potential energy from

                    DPEspring = 1/2kx2 - 1/2kx02

                                 = 1/2(900N/m)(0.25m)2 - 0

                                 = 28J.