A 126.2 N carton is pulled up a frictionless baggage ramp inclined at 32.0 above

Question

A 126.2 N carton is pulled up a frictionless baggage ramp inclined at 32.0 above the horizontal by a rope exerting a 71.8 N pull parallel to the ramp's surface. The carton travels 5.00 m along the surface of the ramp

Part A

Calculate the work done on the carton by the rope.

Part B

Calculate the work done on the carton by the gravity

part C

Calculate the work done on the carton by the normal force of the ramp.

Part D

What is the net work done on the carton?

Part E

Suppose that the rope is angled at 40.6 above the horizontal, instead of being parallel to the ramp's surface. How much work does the rope do on the carton in this case?

Explanation / Answer

a)
Work = force*distance = 71.8*5m = 359 J
b)
Work = mgh = 126.2N*5*sin32 = (-) 347.95 J
c)
the normal force on the box is perpendicular to its movement, so
no work was done
d)
The answer is probably meant to be zero more or less
(359 - 347.95) J
e)
the force is 40.6 - 32 =8.6° above the incline:
the work done is
W = F*cos5.6°*d = 71.8*cos8.6*5 =- 243.66 J

please refer the solve part.

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