Question

Shiela has just arrived at airport and dragging her suitcase to luggage check-in desk. She pulls on the strap with a force of 290N at an angle 35° to the horizontal displacement at 4.5m to the desk. Determine the work done by shiela on the suitcase.
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Answer 1

Answer:

the work done by the Sheela is " W=F×D" that is force ×displacement = 290× 4.5=29×45=1305joules

Answer 2

Answer:

Shiela does a work of 1069.056J on the suitcase.

Explanation:

When any force is exerted on the body and it undergoes some movement and work is said to be done on the body.

Work is related to force and displacement as given,

W=FdCosθ

As given in the question,

F=290N

d=4.5m

θ=35°

On placing the given values in the work equation,

W=290×4.5×Cos35°

W=290×4.5×0.819

W=1069.056J

Hence, Shiela performs a work of 1069.056J on the suitcase.