Conservation of energy (Finding Velocity)

AI Thread Summary
The discussion revolves around calculating the velocity of a 100kg crate after 2 seconds under a variable force P = 20t N, with an initial velocity of 3 m/s down a plane and a coefficient of kinetic friction of 0.3. Participants explore the relationship between momentum and energy, noting the need to account for the impulse of the force over time. The calculations initially yield conflicting results, prompting clarification on integrating the force and considering friction as a negative impulse. Ultimately, the correct velocity after 2 seconds is determined to be approximately 8.2 m/s, following the proper application of the impulse formula. The conversation emphasizes the importance of accurately integrating time-dependent forces in physics problems.
Sam Fred
Messages
15
Reaction score
0

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 ?
IMG_0871.jpg
 
Physics news on Phys.org
Sam Fred said:
I am having problem finding the distance S ?
Think in terms of momentum instead of energy.
 
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
 
Sam Fred said:
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.
 
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 ??
 
Sam Fred said:
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.
 
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 ?
 
Sam Fred said:
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!
 
Thanks ...
 
Back
Top