Question

A 2.90-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spring is attached to a wall, as shown. The initial height of the block is 0.500 m above the lowest part of the slide and the spring constant is 475 N/m.What is the maximum compression of the spring?The spring sends the block back to the left. How high does the block rise?

Answer 1

Law of conservation of energy:

total mechanical energy :

E = KEi + PEi = 0 + mghi = (0.5) * (9.8)*(2.9) = 14.21 J

Where;

KEi= initial kinetic energy at the top

PEi= initial potential energy at the top

hi= 0.50m

g= 9.8 m/s^2

m= 2.9 kg

At the bottom of the ramp, PE= 0, is all kinetic.

The block compresses the spring when it stops is at max compression.

E = Ec = 1/2 k x^2

Where:

k= spring constant 475 N/m

x= compression of the spring

Solve for x:

x =√2E/k = √ (2*14.21 )/475 = 0.24 m (maximum compression of the spring)

The spring sends the block back to the left.

initial energy = Ei = 14.21J

Final energy = m g hf (potential energy at the top)

hf = maximum height

Hf = E/mg = 14.21/ (2.90*9.8) = 0.5m

.