it is simple to change the force diagram to forces parallel with the hill, and normal to the plane.
Friction is parallel to the plane
Weight has two components:
normal to the plane: 196CosTheta
down the plane: 196SinTheta
Pushing force has two components:
up the plane: 290cosTheta
normal to the plain:290sinTheta
Now sum Forces up the plane:
290CosTheta-196sinTheta-65=20*a so you are correct so far.
Now on work, the work done up the plane: Force*distance or
(290CosTheta-196sinTheta-65)13.5
and work normal to the plane
(196CosTheta+290sinTheta)distance
where distance = 13.5sinTheta
Add the two work components.
For time to top,in the up the ramp equation F=ma, solve for acceleration a.
Then, knowing a,
Vf^2=2*a*13.5 solve for Vf.
then, time=13.5/(Vf/2) (distance/avgvel)
A 20.0-kg crate sits at rest at the bottom of a 13.5 m -long ramp that is inclined at 40 ∘ above the horizontal. A constant horizontal force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.
A) What is the total work done on the crate during its motion from the bottom to the top of the ramp?
B) How much time does it take the crate to travel to the top of the ramp?
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The block's weight is 196N, so the horizontal component is 196sin40. I summed the forces such that:
Fnet = ma
290 - 65 - 196sin40 = ma
99 = ma
Since W = Fd:
W = 99*13.5 = 1336.5J which is incorrect. What am I missing?
For the second one, I assume you set the correct value for work equal to the change in potential gravitational energy plus change in kinetic energy?
1 answer
1 answer