A 115kg mail bag hangs by a vertical rope 3.3m long. A postal worker then displa

Question

A 115kg mail bag hangs by a vertical rope 3.3m long. A postal worker then displaces the bag to a position 1.8msideways from its original position, always keeping the rope taut.

Part A

What horizontal force is necessary to hold the bag in the new position?

Express your answer using two significant figures.

Part B

As the bag is moved to this position, how much work is done by the rope?

Express your answer using two significant figures.

Part C

As the bag is moved to this position, how much work is done by the worker?

Express your answer using two significant figures.

Explanation / Answer

A)

here , for sideways = 1.8 m

angle made by the rope with verticle , theta = arcsin(1.8/3.3)

theta = 33.1 degree

Now , let the tension in the string is T

let the horizontal force is F

T*cos(theta) = mg

T*sin(theta) = F

F/mg = tan(theta)

F = 115 * 9.8 * tan(33.1)

F = 733.44 N

the horizontal force needed is 733.44 N

B)

as string is always taut , tension is always perpensicular to direction of motion ,

hence ,

Work done by string is ZERO

C)

WOrk done by worker = change in potential energy

WOrk done by worker = mg * l * (1 - cos(theta))

WOrk done by worker = 115 * 9.8 * 3.3 * ( 1 - cos(33.1))

WOrk done by worker =603.5 J

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