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Homework Statement
The spring in the figure below is initially compressed by 0.4 m in the position shown. If released from this position, block A travels 0.9 m before coming to a stop. The kinetic coefficient of friction is 0.8. What is the spring constant? (The spring is not fastened to block A)
Homework Equations
Elastic potential energy=0.5kx^2
Work=Fd
The Attempt at a Solution
The force of gravity acting on the block=9.81*4=39.24 N
The force of gravity acting on the block parallel to the ramp=sin(10°)*39.24=6.81 N
The force of gravity acting on the block perpendicular to the ramp=cos(10°)*39.2=38.64 N, which is also what the normal force is equal to.
Force of friction=38.64*0.8=30.91 N
Elastic potential energy=(0.5)(0.4)^2(k)
The block will gain kinetic energy from the elastic potential energy + the gravitational potential energy, and lose an amount equal to that (since it stops) due to friction, so:
elastic potential energy+gravitational potential energy=energy lost to friction
(0.5)(0.4)^2(k)+(6.81)(0.9)=(30.91)(0.9)
k≈271
However, the answer I'm given is k=260. I'm not sure if this is a typo, rounding error, or if I'm doing something wrong. Any help would be much appreciated.