Question

A nut can be opened by a lever of length 0.25 cm by applying a force of 80 N. What should be the length of lever, if a force of 32 N is enough to open the nut?

Answer 1

The work done to open the nut in both cases is same.

∴ Work done by the lever in case (ii) = work done by lever in case (i)

Effort × Effort arm in case (ii) = effort × effort arm in case (i)

32 N × Effort arm in case (ii) = 80 N × 0.25 cm

Effort arm in case (ii) \[= \frac{80 \times 0.25}{32} \]

\[= \frac{20}{32} \]

= 0.625 cm