A 9 kg crate is placed at rest on an incline plane (s=0.8, k=0.4), whose angle i

Question

A 9 kg crate is placed at rest on an incline plane (s=0.8, k=0.4), whose angle is 60o. It then slides 4 m down the incline.

a.) Find the work done by each force on the crate after sliding this distance:
work done by normal force: ? J
work done by gravity: ?J
work done by friction force: ?J

b.) Using your results from part a, find the speed of the crate after sliding this distance.
?m/s

c.) Using the work approach only, could you find the time it takes to slide this distance? Discuss the limitations of using the work/energy method vs. using the forces method to solve problems.

Explanation / Answer

Given Mass of the crate , m = 9 kg Coefficient of static friction, s = 0.8 Coefficient of kinetic friction, k = 0.4 Angle of inclination, = 600 Length of the incline , l = 4 m a) Work done by the normal force is , Wn = N *l = 0 J       Work done by gravity is , Wg = mgsin60 *l = 9kg *9.8 m/s^2*0.866 *4 m                                                  = 305.5 J       Work done by frictional force is , Wf = - k mg cos60 * l                                                            = - 0.4 *9kg *9.8 m/s^2*0.5 *4 m                                                       Wf = -70.56 J b) By work energy theorem     Net work done = KEf - KEi     (305.5 J - 70.56 J ) = 1/2*9kg *v^2 - 0            v = 7.22m/s is the speed of the crate after sliding the distance c) Since Work done = Force * l              234.94 J = 9kg * a *4 m         [F = ma]                    a =   6.52 m/s^2 But a = v / t          t = 7.22 m/s / 6.52 m/s^2          t = 1.1 s is the time taken for the crate to slide       Length of the incline , l = 4 m a) Work done by the normal force is , Wn = N *l = 0 J       Work done by gravity is , Wg = mgsin60 *l = 9kg *9.8 m/s^2*0.866 *4 m                                                  = 305.5 J       Work done by frictional force is , Wf = - k mg cos60 * l                                                            = - 0.4 *9kg *9.8 m/s^2*0.5 *4 m                                                       Wf = -70.56 J b) By work energy theorem     Net work done = KEf - KEi     (305.5 J - 70.56 J ) = 1/2*9kg *v^2 - 0            v = 7.22m/s is the speed of the crate after sliding the distance c) Since Work done = Force * l              234.94 J = 9kg * a *4 m         [F = ma]                    a =   6.52 m/s^2 But a = v / t          t = 7.22 m/s / 6.52 m/s^2          t = 1.1 s is the time taken for the crate to slide      

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