# WORK by Constant Force Argh help

1. Oct 15, 2006

### Kalie

WORK by Constant Force....Argh....help

A freight company uses a compressed spring to shoot 1.80 kg packages up a 1.0-m-high frictionless ramp into a truck, as the figure shows. The spring constant is 337 N/m and the spring is compressed 34.0 cm.

A) What is the speed of the package when it reaches the truck?

B) A careless worker spills his soda on the ramp. This creates a 50-cm-long sticky spot with a coefficient of kinetic friction 0.30. Will the next package make it into the truck? Yes or No?

Okay....For Part A this is what I did but it was wrong

First I said F_s=-k x= 337(.34) = 114.38
with this I said that 114.38=1/2mv^2 v=11.28
The F_w=1/2mv^2=ma=1.8*-9.8=-17.64
W=F_w*D=F_w=17.64
W= the change in KE
KE_f=W+KE_i
KE_f=96.74
1/2mv^2=96.74
so v=10.36
That is wrong and I don't know how to get the right answer could someone help?

Part B will probably make more sense after I solve the first part but could someone help me with the setup?

2. Oct 15, 2006

### Staff: Mentor

Here you calculate the force of the spring when it is at max compression.
Then you set this force equal to the KE! :yuck: (Force and energy are two different things!)
Try this: Consider conservation of energy. What's the energy when the spring is fully compressed and the package is at it's lowest point? (You didn't supply the figure, so it's not clear how the spring and ramp are connected. Is the spring on the ramp? Post the figure if you can.)

3. Oct 15, 2006

### Kalie

4. Oct 15, 2006

### Staff: Mentor

OK. Now use conservation of energy to find out how much KE the package has when it gets to the top of the ramp. Hint: Consider spring PE, gravitational PE, and KE.