posted by Anonymous .
A 10 kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20 degrees with the horizontal. If the coefficient of kinetic friction is .40 and the crate is pulled a distance of 5 m, (a) how much work is done by gravitational force? (b) How much work is done by the 100 N force? (c) What is the change in kinetic energy of the crate? (d) What is the speed of the crate after it is pulled 5m?
Before you start answering the various parts of the problem, it is a good idea to draw a free body diagram of forces acting on the block.
The force pulling it up the ramp is
The gravity force (weight) downwards is
M g = 10*9.8 = W = 98 N
The normal force appled to the block by the incline is
Fn = M g cos 20 = 92.1 N
The friction force is
Ff = Fn*Uk = 36.8 N (backwards)
Vertical height change
H = 5 sin 20 = 1.71 m
(a) - W*H = ? (it is negative because the object is raised against the gravity force
(b) F * 5m = ?
(c) (F - Ff)* 5 - W*H = ?
(d) =The answer to (c) equals the kinetic energy change. Use that to get the final kinetic energy. That can tell you what the final speed is, using the rule that K.E. = (1/2) M V^2