Question

A 2-kilogram block is dropped from a height of 0.45 meter above an uncompressed spring, as shown. The spring has an elastic constant of 200 newtons per meter and negligible mass. The block strikes the end of the spring and sticks to it.

 

a. Determine the speed of the block at the instant it hits the end of the spring 

b. Determine the force in the spring when the block reaches the equilibrium position 

c. Determine the distance that the spring is compressed at the equilibrium position 

d. Determine the speed of the block at the equilibrium position 

e. Determine the resulting amplitude of the oscillation that ensues 

f. Is the speed of the block a maximum at the equilibrium position, explain. 

g. Determine the period of the simple harmonic motion that ensues

Answer 1

(a) Apply energy conservation from top to end of spring using h=0 as end of spring.

U = K 

mgh = ½mv²

(9.8)(0.45) = ½ v² 

v = 3m/s

(b) At equilibrium the forces are balanced 

F_net = 0 

F_sp = mg =(2)(9.8) = 19.6N

(c) Using the force from part b, 

F_sp = k ∆x 

19.6 = 200∆x 

∆x = 0.098 m

(d) Apply energy conservation using the equilibrium position as h = 0. 

(Note that the height at the top position is now increased by the amount of ∆x found in part c) 

h_new = h + ∆x = 0.45 + 0.098 = 0.548m 

U_top = U_sp + K _{(at equil)} 

mgh_new = ½ k ∆x² + ½ mv² 

(2)(9.8)(0.548) = ½(200)(0.098)² + ½ (2)(v²) 

v = 3.13m/s

(e) Use the turn horizontal trick. Set equilibrium position as zero spring energy then solve it as a horizontal problem where 

K_equil = U_{sp(at max amp.)} 

½ mv² = ½ k∆x² ½ (2)(3.13)² = ½ (200)(A²) 

A = 0.313m

(f) This is the maximum speed because this was the point when the spring force and weight were equal to each other and the acceleration was zero. Past this point, the spring force will increase above the value of gravity causing an upwards acceleration which will slow the box down until it reaches its maximum compression and stops momentarily.