Work, Energy and Power

Given :

Mass of the block, $ m $ = 0.25 kg

From the work-energy theorem, the potential energy of the spring equals the potential energy of the block :

$ \frac{1}{2} k x^2 = m g x $

Where :

$ k $ is the force constant of the spring

$ x $ is the compression in the spring

Solving for $ kx $ we get :

$ kx = 2mg $

Substituting $ m = 0.25 \, \text{kg} $ :

$ kx = 2 \times 0.25 \times g $

$ kx = 0.5g $

The force applied by the spring on the block is :

$ F = kx \Rightarrow F = 0.5g $

From the free body diagram :

Thus, the maximum force $ N $ exerted by the system on the floor is :

$ N = F + 2g \\ N = 0.5g + 2g \\ N = 2.5g \\ N = 25 \, \text{N} $

Free body diagram of block is shown below.

Now, from the energy conservation,

$\begin{aligned} w & =\Delta K \\\\ \Rightarrow \quad w_F+w_{s p} & =\frac{1}{2} m v^2-0 \\\\ \therefore \quad F_x-\frac{1}{2} k x^2 & =\frac{1}{2} m v^2 \\\\ v & =\sqrt{\frac{2 F_x-k x^2}{m}} \end{aligned}$

For the first displacement $y=0$

Hence, $F_x=0$ and no work is done. For the second displacement, $F_y=-k a$ and $\Delta y=a$

Work $=F_y \Delta y=-k a^2$

Let at any instant of time, the length $A B$ be $l$. Here angle $\theta$ and length $l$ vary with time, then using Pythagorus theorem in $\triangle A B C$

Differentiating both sides w.r.t. '$t$', we get

$x^2+y^2=l^2$

Differentiating both sides w.r.t. '$t$', we get

or $\quad 2 x \frac{d x}{d t}+2 y \frac{d y}{d t}=2 l \frac{d l}{d t}$

As, there no vertical motion of the block, so

$\begin{aligned} & & \frac{d y}{d t} & =0, \frac{d x}{d t}=v_x \text { and } \frac{d l}{d t}=v \\ & \therefore & 2 x v_x & =2 l v \\ & \text { or } & v_x & =\frac{l}{x} v \\ & \text { or } & v_x & =\frac{v}{\left(\frac{x}{l}\right)}=\frac{v}{\sin \theta} \end{aligned}$

Given, $m=70 \mathrm{~kg}, g=10 \mathrm{~ms}^{-2}, h=12 \mathrm{~m}$ In going up and down once.

Number of k-cals burnt $=\left(m g h+\frac{m g h}{2}\right)$

$\begin{aligned} & =\frac{3}{2} m g h \\ & =\frac{3}{2} \times \frac{70 \times 10 \times 12}{4.2 \times 1000} \\ & =3 \mathrm{k}-\mathrm{cal} \end{aligned}$

Total no. of k-cal to be burnt for loosing $7 \mathrm{~kg}$ of weight $=7 \times 9000=63000 \mathrm{~k}$-cal

$\therefore$ Number of times the person has to go up and down the stairs

$\begin{aligned} & =\frac{63000}{3}=21000 \\ & =21 \times 10^3 \text { times } \end{aligned}$