A block of mass 4 kg is compressing a spring with K=2500 N/m a distance of 0.3 m

Question

A block of mass 4 kg is compressing a spring with K=2500 N/m a distance of 0.3 m. The block is then released and moves across a frictionless ramp. The block then hits a ramp angled 20 degrees above horizontal.

a. What is the velocity of the block after it has left the spring but before it hits the ramp?
b.How much work does the spring do on the block?
c. How far up the ramp does the block go?
d.How much work does the ramp do on the block?
e.How much work does gravity do as the block slides up the ramp?

Explanation / Answer

Part A)

From the conservation of energy we have

PE = KE

.5kx2 = .5mv2

v = (kx2/m) = [(2500)(.3)2/(4)] = 7.5 m/s

Part B)

By the work energy theorem

W = KE = .5mv2f - .5mv2i = (.5)(4)(7.5)2 - 0 = 112.5 J

Part C)

From the conservation of energy, KE = PE

.5mv2 = mgh

h = (.5v2)/g = (.5)(7.5)2/(9.8) =   2.87 m high

Since the height is the opposite side of the triangle formed by the 20o ramp, the distance along the ramp is the hypotenuse

Then using sin = O/H, we can solve for H

H = O/sin = (2.87)/(sin 20)   H = 8.39 m

Part D)

The work done by the ramp is also by the work-energy theroem

W = KE

W = KE = .5mv2f - .5mv2i = 0 - (.5)(4)(7.5)2 = -112.5 J (The negative is due to slowing the block down)

Part E)

For gravity, W = PE

W = mgh

W = (4)(-9.8)(2.87) = -112.5 J (The negative indicates that the work is done against the direction of gravity)

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