A spring with a spring constant of 500. N/m is used to propel a 0.500-kg mass up

Question

A spring with a spring constant of 500. N/m is used
to propel a 0.500-kg mass up an inclined plane. The spring
is compressed 30.0 cm from its equilibrium position and
launches the mass from rest across a horizontal surface
and onto the plane. The plane has a length of 4.00 m and is
inclined at 30.0°. Both the plane and the horizontal surface
have a coefficient of kinetic friction with the mass of 0.350.
When the spring is compressed, the mass is 1.50 m from the
bottom of the plane.


b) What is the speed of the mass as it reaches the top of the
plane?
c) What is the total work done by friction from the
beginning to the end of the mass’s motion?

Explanation / Answer

Total work done = Change in Kinetic energy
Kinetic energy initial imparted to the block = .5*500*.09 - (.35*.5*9.8*.3) = 21.98J

Work done by friction when it does not enter the incline = -mga = -.35*.5*9.8*1.2 = -2.058 J

Work done by weight of the body when on the incline = -mgsin.d = -.5*9.8*.5*4 = -9.8 J

Work done by friction of the when the body is on the incline = -mgcos.d = -.35*.5*9.8*.866*4 = -5.94076 J

Total work done = -(5.94076 + 9.8 + 2.058) = -17.8 J

Work done = Change in kinetic energy

-17.8 = KE(final) - 21.98

KE(final) = 4.18 J

a) Velocity at the top be v

so .5mv^2 = 4.18

v = 4.09 m/s

b)Work done by friction = -(2.058+5.941+.515) = -8.5135 J

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