Question

A lever is used to lift a heavy load. When a 52 N force pushes the left end of the lever down 1.28 m, the load rises 0.23m.
4) If the lever is only 84% efficient, what is the work output?
5) If the lever is only 84% efficient, what would be the weight of the load it could lift with this output?

Answer 1

Question#18433

A lever is used to lift a heavy load. When a 52 N force pushes the left end of the lever down 1.28 m, the load rises 0.23 m.

4) If the lever is only 84% efficient, what is the work output?

5) If the lever is only 84% efficient, what would be the weight of the load it could lift with this output?

Solution:

Let:

$$F_1 = 52 \, \text{N}$$ $$S_1 = 1.28 \, \text{m}$$ $$S_2 = 0.23 \, \text{m}$$ $$k = 84\% = 0.84$$ $$A(\text{output}) - ?$$ $$F_2 - ? \text{ output weight}$$

According to the Law of the laver:

$$\frac{F_1}{F_2} = \frac{S_2}{S_1}$$ $$F_2 = \frac{F_1 S_1}{S_2}$$

Such as the laver is only 84% efficient:

$$F_2 = 0.84 \frac{F_1 S_1}{S_2} = 0.84 * \frac{52 \times 1.28}{0.23} = 243.09 \, \text{N}$$ $$A(\text{output}) = F_2 * S_2 = 243.09 * 0.23 = 55.91 \, \text{J}$$

Answer: the output work is 55.91 J, the weight of the load is 243.09 N