A 1kg block is released from rest from the top of an inclined plane. The incline

Question

A 1kg block is released from rest from the top of an inclined plane. The inclined plane has a length L and makes an angle of 40 with the horizontal. The coefficient of kinetic friction between the inclined plane and the block is uk=.20 . At the bottom of the plane there is a frictionless horizontal surface. The block slides ont he horizontal surface until it come to rest after compressing a spring with a spring constant k=100N/m. Please show all Work! Thank You

a. What is the kinetic engergy of the block at the bottom of the plane, right before contacting the spring?

b. How much does the spring compress?

Explanation / Answer

here,

mass of the block , m = 1 kg

theta = 40 degree

uk = 0.2

spring constant , k = 100 N/m

(a)

the kinetic energy of the block at bottom be KE

KE = m*g*( l* sin(40)) - uk*m*g*cos(theta) * l

KE = 1 * 9.8 * l ( sin(40) - 0.2 * cos(40))

KE = 4.8 l J

the kinetic energy is 4.8 l J

(b)

let the compression in the spring be k

0.5 * k * x^2 = KE

0.5 * 100 * x^2 = 4.8*i

x = 0.31 sqrt(l) m

the compression in the spring is 0.31 sqrt(l) m

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