# Homework Help: Energy Equations! Need help with energy of system & coefficient of friction

1. Oct 30, 2008

### EgererI

1. The problem statement, all variables and given/known data

A man with mass 78.0 kg sits on a platform suspended from a movable pulley, and raises himself at constant speed by a rope passing over a fixed pulley. The platform and the pulleys have negligible mass. Assume that there are no friction losses.

Find the increase in the energy of the system when he raises himself 1.10 m. (Answer by calculating the increase in potential energy and also by computing the product of the force on the rope and the length of the rope passing through his hands.)

&

A block with mass m = 0.60 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m (Fig. 7.31). When released, the block moves on a horizontal table top for 1.00 m before coming to rest. The spring constant k is k = 120 N/m. What is the coefficient of kinetic friction, µk between the block and the table?

2. Relevant equations

The big energy equation
KE(translational) + PE(gravitational) + PE(spring) +/- work = KE(translational) + PE(gravitational) + PE(spring) or
1/2mv^2 + mgh + 1/2kx^2 +/- work = 1/2mv^2 + mgh + 1/2kx^2

friction = coefficient-of-friction x normal force

3. The attempt at a solution

for question 1
the work needed to pull the rope so he goes up, I found it to be 254.8 N, since there are three ropes I had to divide the actual work if there were no pullies by three and so that's what I got.
Then I'm just fuzzy on the question and what I'm looking for--I know the answer needs to be in KJ
254.8 N X 1.1m + 78kg X 9.8m/s^2 X 1.1m = 1121.12J
and then converted to KJ = 1.12
that answer is not correct though

for question 2

I figured that the initial part of the Energy equation only needed the potential spring energy and the Work done on it [friction]. The final part of the energy equation I thought to have nothing needed. The distance the block goes is either .8m or 1m

Rearrange the equation and I got:
1/2kx^2 = work
1/2kx^2 = f X d
1/2 X 120n/m X .2^2 = f X .8m
then using f= coefficient X normal force
f=.51
1/2 X 120n/m x .2^2 = f X 1m
f=.408

neither are correct though

2. Oct 31, 2008

### tiny-tim

Welcome to PF!

Hi EgererI! Welcome to PF!
i] there are two ropes … one attached to the platform, the other being pulled by the man (and yeah, I know they're joined )

ii] sorry, but you're completely missing the point …

there are two separate ways of calculating the energy, and the question wants you to practise each of them …

you can use conservation of energy (KE + PE = constant),

or you can use the work-energy theorem (∆W = ∆KE).

Try again!
hmm … you divided by 9.8 x 0.6 … looks right to me.