A boy exerts a force of 210 N on a lever to raise a 1.25 103 N rock a distance of 12 cm. If the efficiency of the lever is 88.7%, how far did the boy move his end of the lever?

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<3,
PhysicsCrazy<3

To find the distance the boy moved his end of the lever, we can use the concept of work and efficiency.

First, let's calculate the work done by the boy on the lever. The formula for work is given by:

Work = Force x Distance

So, the work done by the boy can be calculated as:

Work = 210 N x Distance1

Next, let's calculate the work done on the rock. The formula for work is again:

Work = Force x Distance

So, the work done on the rock can be calculated as:

Work = 1.25 * 10^3 N x 12 cm (Note: we need to convert cm to meters)

We can then set up an equation to represent the efficiency of the lever. The formula for efficiency is given by:

Efficiency = (useful work obtained / total work input) x 100%

Since the efficiency is given as 88.7%, we can write the equation as:

0.887 = (Work done on the rock / Work done by the boy) x 100%

Now, let's substitute the values we have into the equation:

0.887 = (1.25 * 10^3 N x 12 cm) / (210 N x Distance1) x 100%

To solve for Distance1, we can rearrange the equation:

Distance1 = (1.25 * 10^3 N x 12 cm) / (210 N x 0.887)

Now, let's calculate Distance1:

Distance1 = (1.25 * 10^3 N x 12 cm) / (210 N x 0.887)

(Note: Remember to convert centimeters to meters by dividing by 100.)

Distance1 = (1.25 * 10^3 N x 0.12 m) / (210 N x 0.887)

Distance1 = 0.15 m

Therefore, the boy moved his end of the lever a distance of 0.15 meters.

Work done on the rock will be 87% of the work performed by the boy.

210N*X = 0.87*1250N*0.12m
Solve for X