Block striking a horizontal spring

TraceBusta

A moving 8.07 kg block collides with a horizonal spring whose spring constant is 372 N/m. The block compresses the spring a maximum distance of 6.87 cm from its rest position. The coefficient of kinetic friction between the block and the horizontal surface is 0.752.
(a) How much work is done by the spring in bringing the block to rest?
(b) How much mechanical work is done by the force of friction while the block is being brought to rest by the spring?
c) What was the speed of the block when it hit the spring?

for this problem I don't know whether I should start with (a) or not, because the information from the other parts might be useful. I tried to solve (a) but I get the wrong answer.
This is what I did. 1/2 kx^2=.5(372 N/m)(.0687)^2=0.8779 J. <--That is the wrong answer.

TraceBusta

bump...anyone?

I've tried to solve for c) by doing 1/2 kx^2=1/2mv^2 and solve for v=.4664 m/s but that is wrong.

Pyrrhus

While the block is sliding in contact of the spring, it's true the spring will bring it to rest, but there's also friction opposing the block's movement.

Thinking about it... solve it using

[tex] \Delta E = W_{f} [/tex]

Thinking about again... your answer for a looks right to me , what's the books answer?

TraceBusta

with deltaE do you mean change from Kinetic energy to potential energy of the spring?

Pyrrhus

By [tex] \Delta E [/tex] i mean change of mechanical energy, this is not a conservative system you cannot apply conservation of mechanical energy.

[tex] E - E_{o} = W_{f} [/tex]

I will try E when the block stops, and there's only potential spring energy, and of course then the block hits the spring, there's only kinetic energy. Are you using an app like Webassign? try putting a - in your answer.

TraceBusta

for part a i put a - in the answer and it was right..thnks.

i'm still confused as to continue on with parts b) and c)

Pyrrhus

Heh, i know the feeling, i hate those apps too

Pyrrhus

With the equation i gave you, i practically solved the problem for you.

Find the friction force magnitude and use it in the definition of work, you know the radius vector magnitude.

TraceBusta

i think i could solve b) if i knew the answer to c) because then I could use 1/2kx^2-1/2mv^2=(b)?

TraceBusta

Ok. I did Wf=(mu)mg*.0687m and got 4.0899J which is wrong.
/edit I put a - sign and it was right. edit/

Pyrrhus

Let's go back to the definition of Work

[tex] dW = \vec{F} \cdot d \vec{r} [/tex]

so

the result for W will be [tex] |\vec{F}||\vec{r}|cos\theta [/tex]

where [tex] \theta [/tex] is the angle between them

What si the force you want to find its work? Friction force, use the definition for magnitude of friction force

[tex] F_{f} = \mu N [/tex]

You already know the magnitude of the radius vector, and you can know the angle between them.

Pyrrhus

Let me explain you the minus sign... the angle between the radius vector and the force vector is 180 degrees and cosine of 180 degrees is -1, now you know why.

TraceBusta

Awesome, thanks for your help. I figured out c) by doing -(a)-(b)=1/2 mv^2 and solved for v=1.11 m/s

Pyrrhus

It's was a pleasure to be of assistance,