Chapter 10, Problem 39P

To determine

Find the average reaction force for the corresponding time of contact on cushion material.

Answer to Problem 39P

The average reaction force for the corresponding time of contact is calculated and tabulated in Table 1.

Explanation of Solution

Given data:

Weight of the laptop $\left(w\right)$ is 22 N.

Laptop is dropped at a height $\left(h\right)$ is 1m.

Formula used:

Formula to determine the average force is,

$F_{avg}=\frac{m{V}_{\text{f}}-m{V}_{\text{i}}}{\Delta t}$ (1)

Here,

$\Delta t$ is the time period.

m is the mass of the object.

$V_{\text{f}}$ is the final velocity of the object.

$V_{\text{i}}$ is the initial velocity of the object.

Formula to determine the initial velocity of the laptop right before the floor is,

$V_i=\sqrt{2gh}$ (2)

Here,

g is the acceleration due to gravity.

h is the height at which laptop is dropped.

Formula to determine the mass of the laptop is,

$m=\frac{w}{g}$ (3)

Here,

w is the weight.

g is the acceleration due to gravity.

Calculation:

The cushion material reduces the velocity of the laptop to a final velocity of zero. Therefore,

$V_f=0$

Substitute $9.81\frac{\text{m}}{{\text{s}}^{\text{2}}}$ for g and 1 m for h in equation (2).

$\begin{array}{c}{V}_i=\sqrt{2\times 9.81\frac{\text{m}}{{\text{s}}^2}\times 1\text{m}}\\ =\sqrt{19.62\frac{{\text{m}}^2}{{\text{s}}^2}}\\ =4.43\frac{\text{m}}{\text{s}}\end{array}$

Substitute 22 N for w and $9.81\frac{\text{m}}{{\text{s}}^{\text{2}}}$ for g in equation (3).

$\begin{array}{c}m=\frac{22\text{N}}{9.81\frac{\text{m}}{{\text{s}}^{\text{2}}}}\left[\therefore 1\text{N}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =2.24\frac{\text{N}\cdot {\text{s}}^{\text{2}}}{\text{m}}\left[\therefore 1\text{kg}=\frac{\text{N}\cdot {\text{s}}^2}{\text{m}}\right]\\ =2.24\text{kg}\end{array}$

Substitute 0 for $V_f$, 2.24 kg for m, $4.43\frac{\text{m}}{\text{s}}$ for $V_i$, and 0.01 s for $\Delta t$ in equation (1).

$\begin{array}{c}{F}_{avg}=\frac{\left(2.24\text{kg}\times \text{0}\right)-\left(2.24\text{kg}\times 4.43\frac{\text{m}}{\text{s}}\right)}{\left(0.01\text{s}\right)}\\ =\frac{\left(0\right)-\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)}{\left(0.01\text{s}\right)}\\ =\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)\left(\frac{1}{0.01\text{s}}\right)\left[\therefore 1\text{N}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =992\text{N}\end{array}$

Substitute 0 for $V_f$, 2.24 kg for m, $4.43\frac{\text{m}}{\text{s}}$ for $V_i$, and 0.05 s for $\Delta t$ in equation (1).

$F_{avg}=\frac{\left(2.24\text{kg}\times \text{0}\right)-\left(2.24\text{kg}\times 4.43\frac{\text{m}}{\text{s}}\right)}{\left(0.05\text{s}\right)}$

Reduce the equation as follows,

$\begin{array}{c}{F}_{avg}=\frac{\left(0\right)-\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)}{\left(0.05\text{s}\right)}\\ =9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\times \frac{1}{0.05\text{s}}\left[\therefore 1\text{N}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =198\text{N}\end{array}$

Substitute 0 for $V_f$, 2.24 kg for m, $4.43\frac{\text{m}}{\text{s}}$ for $V_i$, and 0.1 s for $\Delta t$ in equation (1).

$\begin{array}{c}{F}_{avg}=\frac{\left(2.24\text{kg}\times \text{0}\right)-\left(2.24\text{kg}\times 4.43\frac{\text{m}}{\text{s}}\right)}{\left(0.1\text{s}\right)}\\ =\frac{\left(0\right)-\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)}{\left(0.1\text{s}\right)}\\ =\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)\left(\frac{1}{0.1\text{s}}\right)\left[\therefore 1\text{N}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =99.2\text{N}\end{array}$

Substitute 0 for $V_f$, 2.24 kg for m, $4.43\frac{\text{m}}{\text{s}}$ for $V_i$, and 1 s for $\Delta t$ in equation (1).

$\begin{array}{c}{F}_{avg}=\frac{\left(2.24\text{kg}\times \text{0}\right)-\left(2.24\text{kg}\times 4.43\frac{\text{m}}{\text{s}}\right)}{\left(1\text{s}\right)}\\ =\frac{\left(0\right)-\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)}{\left(1\text{s}\right)}\\ =\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)\left(\frac{1}{1\text{s}}\right)\left[\therefore 1\text{N}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =9.92\text{N}\end{array}$

Substitute 0 for $V_f$, 2.24 kg for m, $4.43\frac{\text{m}}{\text{s}}$ for $V_i$, and 2 s for $\Delta t$ in equation (1).

$\begin{array}{c}{F}_{avg}=\frac{\left(2.24\text{kg}\times \text{0}\right)-\left(2.24\text{kg}\times 4.43\frac{\text{m}}{\text{s}}\right)}{\left(2\text{s}\right)}\\ =\frac{\left(0\right)-\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)}{\left(2\text{s}\right)}\\ =\left(9.92\frac{\text{kg}\cdot \text{m}}{\text{s}}\right)\left(\frac{1}{2\text{s}}\right)\left[\therefore 1\text{N}=\frac{\text{kg}\cdot \text{m}}{{\text{s}}^{\text{2}}}\right]\\ =4.96\text{N}\end{array}$

Thus, the average reaction force for the corresponding time of contact is calculated and tabulated in table 1.

Table 1

Time of contact (Seconds)	The average reaction force (N)
0.01	992N
0.05	198N
0.1	99.2 N
1.0	9.92 N
2.0	4.96 N

Conclusion:

Hence, the average reaction force for the corresponding time of contact is calculated and tabulated.