Question

a boy exerts a force on a lever to raise a 1250 n rock a distance of 0.13 m. the boy moved his end of the lever 0.722 m. what is the exerted force on the lever?
a 142 n
b 123 n
c 225 n
d 325 n

Explanation:

Step1: Recall the principle of work

Work done by the boy ($W_{boy}$) should equal the work done on the rock ($W_{rock}$) (assuming ideal lever, no energy loss). The formula for work is $W = F \times d$, where $F$ is force and $d$ is distance.

Step2: Set up the equation

Let $F_{boy}$ be the force exerted by the boy, $F_{rock} = 1250\ N$, $d_{boy} = 0.722\ m$, $d_{rock} = 0.13\ m$. So $W_{boy}=W_{rock}\implies F_{boy}\times d_{boy}=F_{rock}\times d_{rock}$.

Step3: Solve for $F_{boy}$

Rearrange the formula: $F_{boy}=\frac{F_{rock}\times d_{rock}}{d_{boy}}$. Substitute the values: $F_{boy}=\frac{1250\ N\times0.13\ m}{0.722\ m}$. Calculate the numerator: $1250\times0.13 = 162.5$. Then divide by $0.722$: $\frac{162.5}{0.722}\approx225\ N$.

Answer:

c. 225 N