- #1

CollegeStudent

- 109

- 0

## Homework Statement

A low friction cart has a spring plunger. The plunger and spring are compressed 6.80 cm and locked in place. The cart is launched from rest up an inclined track tipped at an angle of 13.5°. When the spring is released the cart travels 76.4 cm up the incline from its start position before rolling back down. (a) Assuming frictionless conditions, what is the spring constant of the spring if the cart has a mass of 246 g? (b) The cart rolls back down and is stopped by the spring. Will the spring compress more than, less than, or equal to the original compression of 6.80 cm. Give reasons for your answer. (c) Describe all the energy transformations that take place from start to finish.

## Homework Equations

W = ΔK + ΔU_g + ΔU_s

K = F/Δx

1/2mv²

1/2kx²

## The Attempt at a Solution

(a) Assuming frictionless conditions, what is the spring constant of the spring if the cart has a mass of 246 g?

So I started by using the work equation...we're assuming this system is frictionless so W = 0

there is no kinetic energy in this one so

1/2kx² = mgh

Since I'm solving for k I rearranged this like

k = (2mgh)/x²

the thing is...when I solved for h (height) I did the .764m * sin(13.5) ...but wouldn't that be what "x" is as well?

I know that when using

K = F / Δx ...this Δx would be the amount the spring is compressed here...but in that other equation I think it's the distance traveled

I'm just getting really confused here

any hints?