- #1
Brainsplosion
- 2
- 0
You are designing a delivery ramp for crates containing exercise equipment. The crates weighing F_1 will move at a speed of v at the top of a ramp that slopes downward at an angle \phi. The ramp exerts a kinetic friction force of F_2 on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of L along the ramp. Once stopped, a crate must not rebound back up the ramp.
Calculate the force constant of the spring that will be needed in order to meet the design criteria.
http://img269.imageshack.us/img269/1397/asdfbjl.png
This is what I've tried so far:
initial energy: (Lsin\phi)F_1+1/2*mv^2
final energy: 1/2*kx^2
lost energy: F_2L
initial= final + lost
(Lsin\phi)F_1+1/2*mv^2 = 1/2*kx^2 + F_2L (we shall call this equation 1)
From the free body diagram of the crate resting on the spring at the bottom of the ramp:
kx=F_1sin\phi+F_2 (we shall call this equation 2)
I'm not quite sure all of the above is correct, but if so.. I don't think there's supposed to be the variable of m in there. Should I substitute F_1/g ?
Then, am I supposed to solve for x in equation 2 then plug that into equation 1?
Calculate the force constant of the spring that will be needed in order to meet the design criteria.
http://img269.imageshack.us/img269/1397/asdfbjl.png
This is what I've tried so far:
initial energy: (Lsin\phi)F_1+1/2*mv^2
final energy: 1/2*kx^2
lost energy: F_2L
initial= final + lost
(Lsin\phi)F_1+1/2*mv^2 = 1/2*kx^2 + F_2L (we shall call this equation 1)
From the free body diagram of the crate resting on the spring at the bottom of the ramp:
kx=F_1sin\phi+F_2 (we shall call this equation 2)
I'm not quite sure all of the above is correct, but if so.. I don't think there's supposed to be the variable of m in there. Should I substitute F_1/g ?
Then, am I supposed to solve for x in equation 2 then plug that into equation 1?
Last edited by a moderator: