A luggage handler pulls a suitcase of mass 18.9-kg up a ramp inclined at an angl

Question

A luggage handler pulls a suitcase of mass 18.9-kg up a ramp inclined at an angle 27.0 degrees above the horizontal by a force F of magnitude 149-N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.260. The suitcase travels a distance 4.20-m along the ramp.

A.) Calculate the work done on the suitcase by the force F.
W(F) = 149 x 4.2 = 625.8 626J

B.) Caclulate the work done on the suitcase by the gravitational force.

C.) Calculate the work done on the suitcase by the normal force.

D.) Calculate the work done on the suitcase by the friction force.

W(f) = u*N = umgcos(27)d = (.26)(18.9)(-9.8)cos(27)(4.2) = -180J

E.) Calculate the total work done on the suitcase.

F.) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 4.2 m along the ramp?

Explanation / Answer

Given that The mass of suitcase, m = 18.9 kg Angle of inclination, = 27.0 degree Force along the ramp, F = 149 N The coefficient of kinetic friction between the ramp and the incline, = 0.26 The distance traveled by the suitcase, L = 4.20 m (B) The work done by the graitational force is given by the formula               W(g) = L * mg sin                     = (4.20 m)(18.9 kg)(9.8 m/s2) sin27                     = 353 J (C) Normal force is balanced with the one of the components of the weight of the suitcase.                       N = mg cos Since it is balanced force, it will not work on the suitcase. So the work done by the normal force is zero. (E) Total workdone on the suitcase is given by the sum of above all works.          W = W(F) + W(g) + W(f)          W = 626 + 353 - 180          W = 799 J (F) Since the suitcase is initially at rest, its initial kinetic energy is zero.
According to the work energy theorm we have, the change in kinetic energy is equal to the work done.                 W = KE2 - KE1                 W = KE2 - 0             KE2 = 799
       0.5mv2 = 799   By solving the above we get                v = ( 799 / 0.5m)                v = [799 / (0.5*18.9)]                v = 9.19 m/s          (F) Since the suitcase is initially at rest, its initial kinetic energy is zero.
According to the work energy theorm we have, the change in kinetic energy is equal to the work done.                 W = KE2 - KE1                 W = KE2 - 0             KE2 = 799
       0.5mv2 = 799   By solving the above we get                v = ( 799 / 0.5m)                v = [799 / (0.5*18.9)]                v = 9.19 m/s                v = 9.19 m/s         

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