- #1
kieran89
- 2
- 0
1. A crate of mass 10kg is accelerated up a rough incline with an initial speed of 1.5m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20deg with the horizontal. The coefficient of kinetic friction is 0.40 and the crate is pulled a distance of 5.0m. Use the work-energy princple to calculate the speed of the crate after 5m.
Am I right or wrong? 2. KE1 + PE1 + Ffr*d = KE2 + PE2
Coefficient of Friction = Ffr/FnI got 36.83N for the force of friction.
From my working out I know I have the work done by friction but if i put it on the other side I will have a negative number in my final solution.
(therefore my equation below doesn't actually make sense to me, if I am understand correctly? I am saying with that equation that the inital energy includes friction energy!)
3.
KE1 + PE1 + Ffr*d = KE2 + PE2
=1/2 X 10 X 2.25 + 0 + 36.83 X 5 = 1/2 X 10 X v2 + 10 X 9.8 X 5sin20
=11.25 + 184.15 = 5v2 + 167.59
v2 =(195.4 – 167.59) / 5
v = sqrt(5.562)
= 2.538
v= 2.5 m s-1
Am I right or wrong? 2. KE1 + PE1 + Ffr*d = KE2 + PE2
Coefficient of Friction = Ffr/FnI got 36.83N for the force of friction.
From my working out I know I have the work done by friction but if i put it on the other side I will have a negative number in my final solution.
(therefore my equation below doesn't actually make sense to me, if I am understand correctly? I am saying with that equation that the inital energy includes friction energy!)
3.
KE1 + PE1 + Ffr*d = KE2 + PE2
=1/2 X 10 X 2.25 + 0 + 36.83 X 5 = 1/2 X 10 X v2 + 10 X 9.8 X 5sin20
=11.25 + 184.15 = 5v2 + 167.59
v2 =(195.4 – 167.59) / 5
v = sqrt(5.562)
= 2.538
v= 2.5 m s-1