# Conservation of energy (Finding Velocity)

1. Jun 1, 2013

### Sam Fred

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

The 100kg crate shown in Fig. 1 is acted upon by a force having a variable magnitude
P = 20t N, where t is in seconds. Determine the crates’ velocity 2 seconds after P has
been applied. The initial velocity is v1=3m/s down the plane, and the coefficient of
kinetic friction between the crate and the plane is μk = 0.3. (60 points)
Hint) For easy calculation, the gravitational acceleration (g) is assumed 10m/s2
2. Relevant equations
T1 + U1-2 = T2
T kinetic energy
U potential Energy
3. The attempt at a solution
I am having problem finding the distance S ???

2. Jun 1, 2013

### Staff: Mentor

Think in terms of momentum instead of energy.

3. Jun 1, 2013

### Sam Fred

Aha ....
mv1 + P t (dlta t) + mg sin 30 (dlta t) - mg cos 30 Mk (dlta t) = mv2
30 + 80 + 1000 + 519.6 = 100 v2
v2 = 5.9 m/s

4. Jun 1, 2013

### Staff: Mentor

Double check the first two terms.

5. Jun 1, 2013

### Sam Fred

300 + 80 + 1000 - 519.6 = 100 v2
v2 = 8.6 m/s
Shouldn't the friction be negative and what about the P , It is given P = 20 t , shouldn't i multiply it by t=2 twice ??

6. Jun 1, 2013

### Staff: Mentor

Yes.

Since the force P is a function of time, you'll need to integrate.

7. Jun 1, 2013

### Sam Fred

I was wrong about P . P = 20 t → Impulse = ∫ p dt = 20 t2 /2
Thus P = 20 (4) / 2 = 40 N

But isn't the friction impulse negative and thus we get

300 + 40 + 1000 - 519.6 = 100 v2
v2 =8.2 m/s ????

8. Jun 1, 2013

Good!

9. Jun 1, 2013

Thanks ....