Conservation of energy (Finding Velocity)

[Sam Fred]

Homework Statement

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

Homework Equations

T1 + U1-2 = T2
T kinetic energy
U potential Energy

The Attempt at a Solution

I am having problem finding the distance S ?

 

[Doc Al]

I am having problem finding the distance S ?

Think in terms of momentum instead of energy.

 

[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

 

[Doc Al]

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

Double check the first two terms.

 

[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 ??

 

[Doc Al]

Shouldn't the friction be negative

Yes.

and what about the P , It is given P = 20 t , shouldn't i multiply it by t=2 twice ??

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

 

[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 ?

 

[Doc Al]

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 ?

Good!

 

[Sam Fred]

Thanks ...