A 350 g block moving vertically upward collides with a light vertical spring and compresses it 4.50 cm

Question

A 350 g block moving vertically upward collides with a light vertical spring and compresses it 4.50 cm before coming to rest. If the spring constant is 50.0 N/m, what was the initial speed of the block? (Ignore energy losses to sound and other factors during the collision.)

Answer 1

Use the law of conservation of energy in this problem.

Kinetic energy of block = Energy absorbed by spring

KE of block = (1/2)(m)(V^2) --- call this Equation 1

where

m = mass of the block = 350 grams = 0.350 kg.

V = velocity of the block = unknown

Energy absorbed by spring = mgx + (1/2)kx^2 --- call this Equation 2

where

m = 0.350 kg (as defined above)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

x = length at which spring was compresses = 4.50 cm = 0.045 m (given)

k = spring constant = 50 N/m (given)

Since Equation 1 = Equation 2

(1/2)(m)(V^2) = mgx + (1/2)kx^2

and substituting appropriate values,

(1/2)(0.350)V^2 = (0.350)(9.8)(0.045) + (1/2)(50)(0.045^2)

(1/2)(0.350)V^2 = 0.204975

V^2 = 1.171286

V = 1.08 m/sec.