Ramp, block, and elastic potential energy

[fereas]

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.

 

[Steely Dan]

You're very close. Your mistake is in the gravitational potential energy. You are told that it travels 0.9 m along the surface; so is the block initially 0.9m vertically above the final state, or is it a different amount?

 

[fereas]

You're very close. Your mistake is in the gravitational potential energy. You are told that it travels 0.9 m along the surface; so is the block initially 0.9m vertically above the final state, or is it a different amount?

I found the gravitational energy by multiplying the amount of gravitational force acting parallel to the ramp (6.81 N) by the distance it acts for:

≈(6.81)(0.9)
≈6.129 J

Using the gravitational potential energy equation, I get the same answer:

=mgh
=(4)(9.81)(sin(10)*0.9)
≈6.13 J

 

[darkxponent]

Try solving it smply.

Work done on the box = Change in K.E

 

[Steely Dan]

I found the gravitational energy by multiplying the amount of gravitational force acting parallel to the ramp (6.81 N) by the distance it acts for:

≈(6.81)(0.9)
≈6.129 J

Using the gravitational potential energy equation, I get the same answer:

=mgh
=(4)(9.81)(sin(10)*0.9)
≈6.13 J

My apologies. I saw no sin(10) in the last line and assumed you had overlooked it. The work looks correct to me otherwise.