91Ó°ÊÓ

The given data can be listed below as-

• The value of the Young’s modulus is$Y=2×{10}^{11}N/m^2$ .
• The length of the steel is$L=0.28m$ .
• The mass of the object is $m=85kg$.

The Young’s modulus is described as the ratio of the stress of an object to the strain of that object.

The equation of the Young’s modulus gives the magnitude of the normal force and the compression of the steel cube.

(a)

The equation of the magnitude of the normal force can be expressed as-

$F_N=mg$

Here, $F_N$ is the normal force and m and g are the mass and the acceleration due to gravity.

Substituting the values in the above equation,

$$\begin{gathered} F_N=85kg×9.81m/s^2 \\ =833.85N \end{gathered}$$

Thus, the magnitude of the normal force is $833.85N$.

b)

the equation of the Young’s modulus of a wire can be expressed as-

$Y=\frac{FlA}{\left(∆LlL\right)}$

Here, Y is the Young’s modulus of a wire, F is the normal force, A is the area and$\left(∆L|L\right)$ is the change in the length.

The above equation can also be written as,

$$\begin{gathered} Y=\frac{\left(F|A\right)}{\left(∆L|L\right)} \\ =\frac{FL}{∆LA} \\ =\frac{FL}{∆L\left(L^2\right)}\left[A=L^2\right] \\ =\frac{F}{∆LL} \end{gathered}$$

Hence, further as,

$∆L=\frac{F}{YL}$

Substituting the values in the above equation,

$∆L=\frac{833.85N}{\left(2×{{10}^{11}}^{N/m^2}\right)\left(0.28m\right)}$

Thus, the compression of the steel cube is $1.48×{10}^{-8}m$.