The maximum force that can be borne by the nut placed in a nut cracker is 200 N. The length of the cracker is 20 cm and the nut is placed at a distance of 15 cm from the free end of the nut cracker. If a boy can apply a maximum force of 25 N, find whether he can crack the nut. If not, find the length of the extension rod that should be attached to the cracker handle so that the boy can crack the nut.

Dear Student

Nut cracker is second class lever.  Total length of cracker is 20 cm that is effort arm. Nut is placed 15 cm from free end that is 5 cm from fulcrum that is load arm

Mechanical advantage in terms of effort arm and load arm can be given as
$$\begin{gathered} MA =\frac{Effort arm}{Load arm} \\ MA =\frac{20 cm}{5 cm} \\ MA =4 \end{gathered}$$

In terms of load and effort mechanical advantage can be given as
$$\begin{gathered} MA=\frac{Load}{Effort} \\ 4=\frac{Load}{25} \\ Load = 100 N \end{gathered}$$
So when boy applies force 25 N , the force applied on load is 100 N which is not able to crack the nut.
So  with 25 N effort and 200 N load MA advantage will be
MA = 200 / 25 = 8

So length of nut cracker or length of effort arm should be
8 = effort arm/ 5
Effort arm = 40 cm
So extended length of rod should be 40 -20 = 20 cm.

Regards