91Ó°ÊÓ

1. The mass of the block of cheese is, $m=0.250\text{kg}$$m=0.250kg$.
2. The mass of the elevator cab is,$M=900\text{kg}$.
3. The cab is moving upward by a cable through a distance,$d_{\text{1}}=2.40\text{m}$.
4. The cab is moving upward by a cable through a distance,$d_{\text{2}}=10.5\text{m}$.
5. For , the normal force on the block from the floor is, $F_{\text{N}}=3.00\text{N}$.
6. For , the work done on the cab by the force from the cable is $92.61×{10}^3\text{J}$

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.Draw the free body diagram anduse the concept of Newton’s second law and work done.

For $d_1$,the work done on the cab by the force from the cable

Figure (a) is the free body diagram of the system of the cab with the cheese block.

According to Newton’s second law,

$F_{\text{net}}=ma$

Substitute the values in the above expression, and we get,

$T+F_{\text{N}}-\left(m+M\right)g=\left(m+M\right)a$ (1)

For the system as cheese, we can write,

$$\begin{gathered} F_{\text{N}}-mg=ma \\ a=\frac{F_{\text{N}}-mg}{m} \end{gathered}$$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}a& =& \frac{\left(3.00\text{N}\right)-\left(0.250\text{kg}×9.8{\text{m/s}}^2\right)}{0.250\text{kg}}\\ a& =& \frac{\left(3.00·\left(1\text{N}×\frac{1\text{kg}·{\text{m/s}}^{\text{2}}}{1\text{N}}\right)\right)-\left(0.250\text{kg}×9.8{\text{m/s}}^2\right)}{0.250\text{kg}}\\ & =& \frac{\left(3.00\text{kg}·\text{m/s}\right)-\left(0.250\text{kg}×9.8{\text{m/s}}^2\right)}{0.250\text{kg}}\\ a& =& 2.20{\text{m/s}}^{\text{2}}\\ & & \end{array}$

The force on the cab by the cable can be calculated as,

$$\begin{gathered} T=\left(m+M\right)a+\left(m+M\right)g-F_{\text{N}} \\ T=\left(m+M\right)\left(a+g\right)-F_{\text{N}} \end{gathered}$$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}T& =& \left(0.250\text{kg}+900\text{kg}\right)\left(2.20\frac{\text{m}}{{\text{s}}^{\text{2}}}+\text{9.8}\frac{\text{m}}{{\text{s}}^{\text{2}}}\right)-3.00\text{N}\\ & =& \left(0.250+900\right)\left(2.20+\text{9.8}\right)·\left(1\text{kg}·{\text{m/s}}^{\text{2}}×\frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\right)-3.00\text{N}\\ & =& \left(0.250+900\right)\left(2.20+\text{9.8}\right)\text{N}-3.00\text{N}\\ T& =& 1.08×{10}^4\text{N}\\ & & \end{array}$

The work done by the cable on the cab is,

$W=T{d}_{\text{1}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} W=1.08×{10}^4\text{N}×2.40\text{m} \\ W=2.59×{10}^4\text{J} \end{gathered}$$

Thus, for $d_{\text{1}}$, the work done on the cab by the force from the cable is$2.59×{10}^4\text{J}$

The work done by the cable on the cab is,

$$\begin{gathered} W=T{d}_{\text{2}} \\ T=\frac{W}{d_{\text{2}}} \end{gathered}$$

From equation (i), the normal force acting on the cheese block is,

$F_{\text{N}}=\left(m+M\right)a+\left(m+M\right)g-T$

Substitute the value Of T in the above expression, and we get,

$F_{\text{N}}=\left(m+M\right)\left(a+g\right)-\left(\frac{W}{d_{\text{2}}}\right)$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}{F}_{\text{N}}& =& \left(0.250\text{kg}+900\text{kg}\right)\left(2.20\frac{\text{m}}{{\text{s}}^{\text{2}}}+9.8\frac{\text{m}}{{\text{s}}^2}\right)-\left(\frac{2.59×{10}^4\text{J}}{10.5\text{m}}\right)\\ & =& \left(0.250+900\right)\left(2.20+9.8\right)·\left(1\text{kg}·{\text{m/s}}^{\text{2}}×\frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\right)-\left(\frac{2.59×{10}^4}{10.5}\right)\left(\frac{1\text{J}}{1\text{m}}×\frac{1\text{N}}{\frac{1\text{J}}{1\text{m}}}\right)\\ & =& \left(0.250+900\right)\left(2.20+9.8\right)\text{N}-\left(\frac{2.59×{10}^4}{10.5}\right)\text{N}\\ F_{\text{N}}& =& 2.45\end{array}$

Thus, for $d_{\text{2}}$, the magnitude of the normal force is $2.45\text{N}$.