# Work done and energy transfer

1. Aug 4, 2015

### Zynoakib

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

2. Relevant equations
KE = 1/2mv^2
f = -kx
U = 1/2kx^2
Kinematic equations
F = ma

3. The attempt at a solution
Horizontal components of the block's weight: 8.38N

Compression of the spring:
f = -kx
8.37 = -500x
x = 0.01674 m , which is wrong and I would be grateful if someone can explain to me why it is wrong

Then, I changed the approach

Find the acceleration of the block

F= ma
8.37 = 2.5a
a = 3.348 ms^-2

u = 0.75, v = ?, a = 3.348, s = 0.3

v^2 = u^2 + 2as
v^2 = 0.75^2 + 2(3.348)(0.3)
v = 1.603 m/s

W = KE change when the blocj hit the spring
W = 1/2(2.5)(1.603)^2 - 1/2(2.5)(0)^2
W = 3.212J

Since work done can also be treated as potential energy

U = 1/2kx^2
3.212 = 1/2(500)x^2
x = 0.113 m

and is still wrong, why???

Thanks!

Last edited: Aug 4, 2015
2. Aug 4, 2015

### DEvens

The block is trading potential energy from height for kinetic energy. Then it hits the spring. It starts trading kinetic energy for spring energy.

When does the block stop? When its kinetic energy reaches zero. That happens when the compression energy of the spring equals the potential energy the block has lost by sliding down the ramp.

Don't forget the block is still sliding down after it contacts the spring, so it still loses height until it stops.