QUIZ 5

  1. A uniform 15-N force constant acts on a 5-kg crate with 37˚ above the horizontal as shown in figure. The crate is originally at rest on the ground. The coefficient of friction between the crate and the ground is 0.25. Use work-energy principle to find the speed of the crate after it moves 50m horizontally.
Solution. From the force diagram we find the friction force exerted on the crate

                 Ffr = mkFN =  mk(mg - Fsinq).

             We find the net work done to the crate from

                 SW = Fdcosq  - Ffrd = [Fcosq - mk(mg - Fsinq)]d.

             From work-energy principle we have

                 SW = DEk = 1/2m(vf2 - v02);

                 [Fcosq - mk(mg - Fsinq)]d = 1/2m(vf2 - v02);

                 {(15N)cos37° - (0.25)[(5kg)(9.80m/s2) -  (15N)sin37°)]}(50m)

                   = 1/2(5kg)(vf2 - 0), which gives vf  = 6.3m/s.

 
2. What is the total energy of the system when the block falls freely from height of h above an unstretched spring and stays at rest with the spring compressed at a distance of x as shown in the figure.
Solution. We find the energy of the system from

              E = PEs + PEg = 1/2kx2 - mg(h + x)