Question

A crate of mass 10.6 kg is pulled up a rough incline with an initial speed of 1.52 m/s. The pulling force is 106 N parallel to the incline, which makes an angle of 19.4 degree with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.06 m. How much work is done by the gravitational force on the crate? J Determine the increase in internal energy of the crate-incline system owing to friction. J How much work is done by the 106-N force on the crate? J What is the change in kinetic energy of the crate? J What is the speed of the crate after being pulled 5.06 m? m/s

Answer 1

(a) Work done by gravitai onal force = (106 N)(5.06m)cos(90+19.4) -178 J First find normal force Alon y-direction N-mg cos 19.4 -0 N mg cos 19.40 (10.6 kg) (9.8m/s2)cos 19.4° = 97.98 N Find frictional force f (0.4)(97.98N) 39.19 N Now, find work done due to fricitional force - (39.19 N) (5.06 m)cos 180 -198J In this case, force and displacem ent are in same direction Work done, w, (106N) (5.06m)cos 0 536 N d) Change in kinetic energy of crate -178 J-198 J0+536N 59.88 J 160. J Change in kinetic energy AK (106)[ヅ-(1 52)2] 159.88(10.6)v,2-(1.52) Solve for v 5.698 m/s Round off the result to 3 significant digits v,-5.70 m/s

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