I assume that the force is acting up along the plane.
Since F = ma, a = 0.4t - 9.8sin20° = 0.4t - 3.35 ft/s^2
v(t) = 0.2t^2 - 3.35t
so, v=0 when t=16.76 s
s(t) = 1/15 t^3 - 1.675t^2
Now just figure how far it has gone (s(16.76)), and how long it will take to get back up the plane.
The 20lb cartis released from rest at time t=0 on the incline surface. The force P=8t lb acts on the cart, where t is the time measured in seconds. A) Determine the distance the cart will move down the inclined surface before reversing direction. B) Find the velocity of the cart when it returns to the point of release.
The angle of inclination is 20 degrees
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