Question

3. An 8.50 kg block is held at a height H1 = 7.50 m. The block is released and lands on a spring whose initial height before the collision is H2 = 3.00 m. The spring has a spring constant of 1.50x103 N/m. (Ignore the size of the block.) a. Use Conservation of Energy to find the speed of the block just before it touches the spring. (5)








b. Find the maximum compression of the spring. (20)



















c. Find the maximum height the block reaches after leaving the spring

Answer 1

a)

mu mgh1 = mgh2 +

8.5 * 2 8.5 +9.8 +7.5 = 8.5*9.8 +3+4=

V = 9.39 m/s

b)

mgh1 = *+mg(h2 - 0)

1.5 + 10 本 8.5 +9.8 +7.5 =. - + 8.5 +9.8 + (3 - ac)

T= 0.76 m

c)

Maximum height the block reaches after leaving the spring = 7.5 m

(Only if the block gets back to its initial height energy will remain conserved.)