A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees abov

Question

A luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 25 degrees above the horizontal by a force F of magnitude 145 N that acts parallel to the ramp.The coefficient of kinetic friction between the ramp and the incline is = 0.30. If the suitcase travels 4.60 m along the ramp, calculate:
A) the work done on the suitcase by force F.
B) the work done on the suitcase by the gravitational force.
C) the work done on the suitcase by the normal force.
D) the work done on the suitcase by the friction force.
E) the total work done on the suitcase.
F) if the velocity of the suitcase is zero at the bottom of the ramp,what is its velocity after it has travelled 4.60 m along the ramp.

Explanation / Answer

A) the work done on the suitcase by force F. W = F d = 145*4.60 = 667 J B) the work done on the suitcase by the gravitational force. W = m g sin(theta) d = 20.0*9.8*sin25 * 4.60 = 381.0326 J C) the work done on the suitcase by the normal force. W = 0 D) the work done on the suitcase by the friction force. W = (mu) m g cos(theta) d = 0.30*20.0*9.8*cos25 * 4.60 = 245.13813 J E) the total work done on the suitcase. W_total = 667- 381.0326 - 245.13813 = 40.82927 J F) if the velocity of the suitcase is zero at the bottom of the ramp,what is its velocity after it has traveled 4.60 m along the ramp. 0.5 m v^2 = W_total 0.5*20.0*v^2 = 40.82927 >>>> v = 2.02 m/s

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