Question

A 10 kg block is placed on an inclined plane that is at an angle of 30 degrees with respect to the horizontal. THe coefficient of kinetic friction between the block and the plane is .1. The block is released from rest at 5m above a spring that is also lying on the plane. The spring has a constant of 50 N/m. What is the max compression of the spring?

Answer 1

Important note - A diagram would have been a great help because I do not get this statement "The block is released from rest at 5m above a spring that is also lying on the plane" ....is this 5 m distance along the incline or height above the ground ???? I will solve for both

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Here, is the method to solve.

First of all, we need to find acceleration of the block

F_net = ma

mgsin - u_kmgcos = ma

a = gsin - u_kcosg

a = 9.8 * sin 30 - 0.1 * 9.8 * cos 30

a = 4.05 m/s²

so,

if 5 m is along the incline , then we can use kinematics to find the final velocity of block as it touches the spring

v² = 2ad

v = sqrt ( 2ad)

v = 6.36 m/s

then, using conservation of energy

1/2mv² = 1/2kx²

10 * 6.36² = 50 * x²

x = 2.846 meters ( I think this is the correct case)

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second case -

If 5 m distance is above the ground

then , we need to use trigonometry

5 / d = sin 30

then d = 10 m

then

v = sqrt ( 2 * 4.05 * 10)

v = 9 m/s

then

x = 4.02 m