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

Answer:

A. Input work = 800 ft.lb

B. Output work = 750 ft.lb

C. Efficiency of block and tackle = 93.75 %

Answer 2

Answer:

A. Input work = 800 ft.lb

B. Output work = 750 ft.lb

C. Efficiency of block and tackle = 93.75 %

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:

$\textrm{Work}=\textrm{Force}\times \textrm{Displacement}$

A.

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

$W_{in}=F\times D_{in}\\W_{in}=200\times 4=800\textrm{ ft.lb}$

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,

$W_{out}=F_{load}\times D_{load}\\W_{out}=600\times 1.25=750\textrm{ ft.lb}$

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 = $\frac{W_{out}}{W_{in}}\times 100=\frac{750}{800}\times 100=93.75\%$

Therefore, efficiency of block and tackle is 93.75 %.