PHYSICS 100A #85014

FALL 2003 QUIZ 4

1. A 2-kg block is rest on the base of 15?incline. A 12-N force exerts on the block in the direction of the incline surface as shown. Find the speed of the block when it is pulled 5m on the incline surface. Assume the effective coefficient of kinetic friction between the block and incline is 0.20.

Solution. We draw free-body diagram as shown. we write SF = ma:

                          x-component: F - mgsinq - Ffr = ma;

                          y-component: FN - mgcosq = 0.

               Combine the two equations and with Ffr = mkFN,  we have

                          F - mgsinq  - mkmgcosq  = ma, 

               which gives 

                         a = [F - mg(sinq  + mkcosq )]/m

                            = {12N - (2kg)(9.80m/s2)[sin15?+ (0.20)cos15°}/(2kg)

                            = 1.57m/s2.  

               We find the speed from the motion equation

                        v2 = v02 + 2aDx;

                        v2= 0 + 2(1.57m/s2)(5m), which gives v = 3.96m/s.

 

  

2. An ultracentrifuge rotor rotates at 50,000 revolutions per minute. The bottom of the test tube is 10.00cm from the axis of rotation as shown. Calculate the centripetal acceleration of the bottom of the test tube.

Solution. We calculate the centripetal acceleration from 

                       aR = v2/r.

               We find v from

                       v = 2pr/T = 2prf = 2(3.14)(0.100m)[(50,000rev/min)/(60s/min)] 

                         = 5.23 x102m/s.

               We find the centripetal acceleration from          

                      aR = v2/r = (5.23 x102m/s)2/(0.1000m) = 2.74 x 106m/s2.