Question

(35 points!!) A block and tackle is used to lift a load weighing 600 lb. The operator lifts the load distance of 1.25 ft by pulling on the rope with a force of 200 lb of a distance of 4 ft.

A. input work = ____ ft*lb
B. Output work = ____ ft*lb
C. efficiency of block and tackle = ____%

Answer 1

A. Input work = 800 ft.lb

B. Output work = 750 ft.lb

C. Efficiency of block and tackle = 93.75 %

Answer 2

A. Input work = 800 ft.lb

B. Output work = 750 ft.lb

C. Efficiency of block and tackle = 93.75 %

Step-by-step explanation:

Given:

Weight of the load,
`W=600` lb.

Distance moved by the load,
`D_(load)=1.25` ft.

Force applied on the rope,
`F = 200` lb

Distance moved by the force on the rope,
`D{in}=4` ft.

Work done by a force causing a displacement in the direction of force is given as:

`\textrm{Work}=\textrm{Force}* \textrm{Displacement}`

A.

Here, Input Work is given by the input force and the displacement caused by the input force. So,

`W_(in)=F* D_(in)\\W_(in)=200* 4=800\textrm{ ft.lb}`

Therefore, the input work is 800 lb.ft

B.

Output Work is given by the output force and the displacement caused by the output force. So,

`W_(out)=F_(load)* D_(load)\\W_(out)=600* 1.25=750\textrm{ ft.lb}`

Therefore, the output work is 750 ft.lb

C.

Efficiency is given as the ratio of Output Work to Input Work expressed as percentage. So,

Efficiency =
`(W_(out))/(W_(in))* 100=(750)/(800)* 100=93.75\%`

Therefore, efficiency of block and tackle is 93.75 %.