91Ó°ÊÓ

Given Data:

• The initial mass of the rocket,$m=125\mathrm{kg}$.
• The constant vertical force of the rocket,$T=1720\mathrm{N}$.
• The weight of the electrical power supply,$w=15.5\mathrm{N}$.

Acceleration of rocket and push on electrical supply:

The acceleration of the rocket and push on the rocket is calculated by using the equilibrium of the rocket in the vertical direction.

(a)

The initial acceleration of the rocket is calculated as:

$T-mg=ma$

Here g is the gravitational acceleration, and its value is $9.8m/s^2,T$ is the constant vertical force for the rocket.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} 1720\mathrm{N}-\left(125\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)=\left(125\mathrm{kg}\right)\mathrm{a} \\ \mathrm{a}=3.96\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the initial acceleration of rocket is $3.96\mathrm{m}/{\mathrm{s}}^2$.

(b)

The initial thrust exerted by floor push is calculated as:

$N=\left(\frac{w}{g}\right)\left(g+a\right)$

Here, N is the initial thrust by floor push, w is the weight of the electrical power supply.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} N=\left(\frac{15.5\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2+3.96\mathrm{m}/{\mathrm{s}}^2\right) \\ N=21.7\mathrm{N} \end{gathered}$$

Therefore, the initial thrust exerted by floor push to electrical supply is $21.7\mathrm{N}$.