Answer: Energy Conservation: Maximum Compression & Max Velocity

[habibclan]

Homework Statement

A 10 kg vox slides 4.0 m down the frictionless ramp shown in the following link. It then collides with the spring whose spring constant is 250 N/m.
a. What is the maximum compression of the spring?
b. At what compression of the spring does the box have its maximum velocity?

http://i196.photobucket.com/albums/aa59/aliatehreem/chapter_10.jpg

Homework Equations

Usf= Ug [ spring potential energy final = gravitational potential energy initial.
0.5 k x^2 = mgy

The Attempt at a Solution

Find height corresponding to 4 m.
h= 4 sin30

a. Then use conservation of energy to find compression.
Usf= Ug [ spring potential energy final = gravitational potential energy initial.
0.5 k x^2 = mgy
0.5 (250) (x ^2) = (10) (9.81) (4sin30)
x= 1.25 m

Apparently, the correct answer is 1.46 m for a. and 19.6 cm for b. Can someone please help me figure out what I am doing wrong? It would be greatly appreciated! I wish that they would give the length of the spring, then I could calculate the gravitational potential energy more accurately.

 

[rl.bhat]

The force acting on the block is mgsin(theta) and distance moved by the block is (4 + x). In solving this problem the length of the spring is not required.

 

[mgb_phys]

Have you taken into account that the pe of the box is minimum when the spring stops not at first contact, so the box slides (4+x) down the slope

For (b) write the energy as KE+PE+spring=0 ( get the sign's correct! ) to find an equation only in V and x.

(Oops posts clashed)

 

[habibclan]

by writinf 4 + x, i get 2.2 m, which is still not the correct answer.

Also, for Ke +PE + spring=0, why is the sum 0?

 

[rl.bhat]

0.5 (250) (x ^2) = (10) (9.81) (sin30)(4+x) .
Solve for x.

 

[habibclan]

0.5 (250) (x ^2) = (10) (9.81) (sin30)(4+x) .
Solve for x.

Nice! Thank you so much!