A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.

Question

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 17.5° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.55 m.
(a) How much work is done by gravity?
(b) How much mechanical energy is lost due to friction?
(c) How much work is done by the 100 N force?
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 5.55 m?

Explanation / Answer

   Mass of crate, m = 10 kg    Initial speed of crate, u = 1.5 m/s    The pulling force, F = 100 N    The angle of inclination, theeta = 17.5 degrees    The coefficient of kinetic friction, µk = 0.4    The distance travelled by the crate, l = 5.55 m    (a) The vertical height reached by the crate, h = l sin(theeta)                                                                         = 5.55 sin(17.5)                                                                         = 1.67 m          The work done by the gravity, W = mgh = 10(9.8)1.67                                                       W = 163.66 J    (b) Mechanical energy lost due to friction = fs = µk mgcos(theeta) l                                                                = 0.4(10)(9.8)(5.55) cos(17.5)                                                                  = 207.5 J    (c) Work done by 100 N force, W = fs = 100(5.55) = 555 J    (d) According to work - energy theorm, the workdone by the force is       equal to the change in kinetic energy of the body             ? The change in kinetic energy of the body = 555 J    (e) Acceleration, a = Force / mass = 100 / 10 = 10 m/s^2          We have an equation of motion, v^2 - u^2 = 2 a s                                                                  v^2 = u^2 + 2 a s                                                                     v^2 = (1.5)^2 + 2(10)(5.55)                                                                        v = 10.6 m/s    Hence the velocity of the crate after being pulled 5.55m is 10.6 m/s                                                                = 207.5 J    (c) Work done by 100 N force, W = fs = 100(5.55) = 555 J    (d) According to work - energy theorm, the workdone by the force is       equal to the change in kinetic energy of the body             ? The change in kinetic energy of the body = 555 J    (e) Acceleration, a = Force / mass = 100 / 10 = 10 m/s^2          We have an equation of motion, v^2 - u^2 = 2 a s                                                                  v^2 = u^2 + 2 a s                                                                     v^2 = (1.5)^2 + 2(10)(5.55)                                                                        v = 10.6 m/s    Hence the velocity of the crate after being pulled 5.55m is 10.6 m/s                                                                  v^2 = (1.5)^2 + 2(10)(5.55)                                                                        v = 10.6 m/s    Hence the velocity of the crate after being pulled 5.55m is 10.6 m/s

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