Physics 212, Quiz #1b: Dec 4, 1997

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Bob finds a spring at rest, hanging from the ceiling. The bottom end of the spring is exactly 1.0 meters below the ceiling. Bob attaches a cannonball of mass 10 kg to the bottom end of the spring.

Question 1: What is the force exerted on the spring by the ball?

   Answer: Force exerted by ball is F = mg downwards

                    F = - (10 kg)*(9.8 m/s^2) = -98 N in the y direction

The spring stretches to a total length of 1.2 meters from the ceiling.

Question 2: What is the spring constant of the spring?

                                            F
   Answer: F = -k*x     therefore    k = - ---
                                            x

           Here F = force exerted BY SPRING = opposite of Force exerted by ball
                                            = +98 N in the y direction

                x = displacement from rest = -0.2 meters in the y direction

           So
                       98 N
                k = - ------ = +490 N/m
                      -0.2 m

Bob removes the ball, but carefully keeps the bottom end of the spring at the same distance from the ceiling. He then attaches a book of mass 2 kg to the bottom end of the spring.

Question 3: When Bob lets go of the spring, what will the acceleration of the book be?

   Answer: The acceleration of the book with the net force on it,
           divided by its mass.  The net force is

               F(net) = F(gravity)          +  F(spring)

                      = -(2 kg)*(9.8 m/s^2) +  98 N    in the y direction

                      = -19.6 N             +  98 N

                      = +78.4 N                        in the y direction

           So

                        F(net)     78.4 N
                    a = ------- = --------- = 39.2 m/s^2   upwards
                          m         2 kg

After Bob lets go, the spring bobs up and down for a while, but eventually settles down and stops moving.

Question 4: How far is the bottom end of the spring from the ceiling now?

   Answer: The force exerted by the book on the spring (downwards)
           will equal the force exerted by the spring on the book (upwards)
           when it is at rest.  So the force exerted by the spring
           will be

                 F(spring) = -F(book) = - (-mg) = mg       upwards
 
                           = (2 kg)*(9.8 m/s^2) = 19.6 N   upwards

           We know the force exerted by the spring is equal to 

                F(spring)  = -kx

           So
                                F(spring)     19.6 N
                       x   = - ---------- = - ------- = - 0.04 m
                                  k           490 N/m
          
           The displacement of the spring from rest is -0.04 meters in
           the y direction.  Since the spring's lower end was originally
           1.0 meters from the ceiling, its final position is

                  position = (1.0 m below ceiling) + (- 0.04 m)

                           = 1.04 m below ceiling

This page maintained by Michael Richmond. Last modified Dec 5, 1997.