A .2 kg block is pushed against a spring with a force constant is 400 N/m compre

Question

A .2 kg block is pushed against a spring with a force constant is 400 N/m compressing it 22 cm. When the block is

released it moves over a horizontal friction free surface for 1.5 m then up an 37o inclined plane. The coefficient of kinetic friction on the incline is .5 (a) What is the velocity of the block when it reaches the foot of the ramp prior to going up the ramp? (b) What is the max height the block obtains? (c) What is the velocity of the block at the foot of the ramp after coming back down the ramp?

Please answer in m/s units. Im stuck where i am going wrong. step by step of the conservation of energy greatly appreciated.

Explanation / Answer

Spring constant k = 400 N/m

Energy stored in spring = 1/2 kx^2 = 1/2 *400*0.22^2 = 9.68 J

a)

There is no energy loss till it reaches foot of incline since it is friction free.

KE = 1/2*mv^2 = 9.68

1/2 *0.2*v^2 = 9.68

v = 9.84 m/s

b)

Suppose it oes distance s on the incline.

Normal force = mg*cos37

Frictio force = 0.5*(mg*cos37)

Friction energy loss = 0.5*mg*cos37*s

PE gain = mg*(s*sin37)

Therefore, 9.68 = 0.2*9.81*s*sin37 + 0.5*0.2*9.81*cos37*s

s = 4.928 m

h = s*sin37 = 2.966 m

c)

PE at top = mg*s*sin37

Friction work = 0.5*mg*s*cos37

Kinetic Energy at bottom = mg*s*sin37 - 0.5*mg*s*cos37

1/2 *mv^2 = mg*s*sin37 - 0.5*mg*s*cos37

1/2*0.2*v^2 = 0.2*9.81*4.928*sin37 - 0.5*0.2*9.81*4.928*cos37

v = 4.425 m/s

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