Chapter 14.3, Problem 1PP

Determine the work of the force when it displaces 2 m.

To determine

The work of a force.

Answer to Problem 1PP

The work done by the force on the block is $600\text{J}$.

Explanation of Solution

Given:

The force acting on the block is $F=500\text{N}$.

The displacement of the block is $s=2\text{m}$.

Draw the free body diagram of block as shown in Figure (a).

Write the formula for work done force $\left(F\right)$.

$U_F=Fs$ (I)

Here, $F$ is the force and $s$ is the displacement of the block.

Conclusion:

Refer Figure (a).

Resolve the force along $x-axis$.

$\begin{array}{c}F=500\text{N}\left(\frac{3}{5}\right)\\ =300\text{N}\end{array}$

Substitute $300\text{N}$ for $F$ and $2\text{m}$ for $s$ in Equation (I).

$\begin{array}{c}{U}_F=300\text{N}\left(2\text{m}\right)\\ =600\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ \text{= 600}\text{J}\end{array}$

Thus, the work done by the force on the block is $600\text{J}$.

To determine

The work of a force.

Answer to Problem 1PP

The work done by the force on the block is $0\text{J}$.

Explanation of Solution

Given:

The force acting on the block is $F=98.1\text{N}$.

The displacement of the block is $s=2\text{m}$.

Draw the free body diagram of block as shown in Figure (b).

Write the formula for work done force $\left(F\right)$.

$U_F=Fs$ (I)

Here, $F$ is the force and $s$ is the displacement of the block.

Conclusion:

Refer Figure (1).

The force acting on the block does not cause any displacement of the block. Hence the work done by the force is zero.

Thus, the work done by the force on the block is $0\text{J}$.

To determine

The work of a force.

Answer to Problem 1PP

The work done by the force on the block is $16\text{J}$.

Explanation of Solution

Given:

The force acting on the block is $F=\left(6s^2\right)\text{N}$.

The displacement of the block is $s=2\text{m}$.

Draw the free body diagram of block as shown in Figure (c).

Write the formula for work done force $\left(F\right)$.

$U_F={\displaystyle \underset{s_1}{\overset{s_2}{\int }}Fds}$ (I)

Here, $F$ is the force and $s_1,s_2$ is the initial and final displacement of the block.

Conclusion:

Refer Figure (c).

Substitute $6s^2$ for $F$, $0$ for $s_1$ and $2\text{m}$ for $s_2$ in Equation (I).

$\begin{array}{c}{U}_F={\displaystyle \underset{0}{\overset{2}{\int }}6s^2}ds\\ =6{\left[\frac{s^3}{3}\right]}_0^2\\ =2\left(2^3-0^3\right)\\ =16\text{J}\end{array}$

Thus, the work done by the force on the block is $16\text{J}$.

To determine

The work of a force.

Answer to Problem 1PP

The work done by the force on the block is $\text{ 120}\text{J}$.

Explanation of Solution

Given:

The force acting on the block is $F=500\text{N}$.

The displacement of the block is $s=2\text{m}$.

Draw the free body diagram of block as shown in Figure (d).

Write the formula for work done force $\left(F\right)$.

$U_F=Fs$ (I)

Here, $F$ is the force and $s$ is the displacement of the block.

Conclusion:

Refer Figure (d).

Resolve the force along $x-axis$.

$\begin{array}{c}F=100\text{N}\left(\frac{3}{5}\right)\\ =60\text{N}\end{array}$

Substitute $60\text{N}$ for $F$ and $2\text{m}$ for $s$ in Equation (I).

$\begin{array}{c}{U}_F=60\text{N}\left(2\text{m}\right)\\ =120\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ \text{= 120}\text{J}\end{array}$

Thus, the work done by the force on the block is $\text{ 120}\text{J}$.

To determine

The work of a force.

Answer to Problem 1PP

The work done by the force on the block is $24\text{J}$.

Explanation of Solution

Given:

The displacement of the block is $s=2\text{m}$.

The given $F-s$ graph is shown in Figure (e).

Draw the free body diagram of block as shown in Figure (1e).

The graph consists of two geometrical cross sectional areas namely triangle and rectangle.

Write the formula for work done force $\left(F\right)$.

$\begin{array}{c}{U}_F=\text{Area under the }F-s\text{}\text{graph }\times \frac{4}{5}\\ =\left(A_r+A_t\right)\times \frac{4}{5}\end{array}$ (I)

Here, $F$ is the force and $s$ is the displacement of the block.

Refer Figure (e).

Write the formula for triangle.

$A_t=\frac{1}{2}{b}_{1}h$ (II)

Write the formula for rectangle.

$A_r=b_{2}h$ (III)

Conclusion:

Refer Figure (e).

Calculate the area under the $F-s$ graph.

Substitute $1\text{m}$ for $b_1$ and $20\text{N}$ for $h$ in Equation (II).

$\begin{array}{c}{A}_t=\frac{1}{2}\left(1\text{m}\right)\left(20\text{N}\right)\\ =10\text{N}\cdot \text{m}\end{array}$

Substitute $1\text{m}$ for $b_2$ and $20\text{N}$ for $h$ in Equation (III).

$\begin{array}{c}{A}_r=\left(1\text{m}\right)\left(20\text{N}\right)\\ =20\text{N}\cdot \text{m}\end{array}$

Calculate the work done by the force $\left(F\right)$.

Substitute $10\text{N}\cdot \text{m}$ for $A_t$ and $20\text{N}\cdot \text{m}$ for $A_r$ in Equation (I).

$\begin{array}{c}{U}_F=\left(10\text{N}\cdot \text{m}+20\text{N}\cdot \text{m}\right)\times \frac{4}{5}\\ =30\times \frac{4}{5}\\ =24\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ =24\text{J}\end{array}$

Thus, the work done by the force on the block is $24\text{J}$.

To determine

The work of a spring force.

Answer to Problem 1PP

The work done by the force on the block is $\text{40}\text{J}$.

Explanation of Solution

Given:

The stiffness of the spring is $10\text{N/m}$.

The spring is originally compressed to $3\text{m}$.

Draw the free body diagram of block as shown in Figure (f).

Write the formula for work done force $\left(F_{sp}\right)$.

$U_{F_{sp}}=\frac{1}{2}k\left(s_0{}^2-s_1{}^2\right)$ (I)

Here, $k$ is the stiffness of the spring, $s_0$ is the unstretched length of the spring and $s_1$ is the final displacement of the spring.

Conclusion:

Refer Figure (f).

Here, the block is subjected to spring force only.

When the spring is originally compressed to $3\text{m}$, the unstretched length of the spring is

$s_0=3\text{m}$.

When the spring is released, the displaced to $2\text{m}$. Hence the final displacement of the spring is $s_1=3\text{m}-2\text{m}=1\text{m}$.

Substitute $10\text{N/m}$ for $k$, $3\text{m}$ for $s_0$ and $1\text{m}$ for $s_1$ in Equation (I).

$\begin{array}{c}{U}_{sp}=\frac{1}{2}\times 10\text{N/m}\left[{\left(3\text{m}\right)}^2-{\left(1\text{m}\right)}^2\right]\\ =5\times 8\\ \text{= 40}\text{J}\end{array}$

Thus, the work done by the spring force on the block is $\text{40}\text{J}$.

To determine

The work of a force.

Answer to Problem 1PP

The work done by the force on the block is $-\text{160}\text{J}$.

Explanation of Solution

Given:

The force acting on the block is $F=100\text{N}$.

The displacement of the block is $s=2\text{m}$.

Draw the free body diagram of block as shown in Figure (g).

Write the formula for work done force $\left(F\right)$.

$U_F=Fs$ (I)

Here, $F$ is the force and $s$ is the displacement of the block.

Conclusion:

Refer Figure (g).

Resolve the force along $x-axis$.

$\begin{array}{c}F=-100\text{N}\left(\frac{4}{5}\right)\\ =-80\text{N}\end{array}$

Substitute $80\text{N}$ for $F$ and $2\text{m}$ for $s$ in Equation (I).

$\begin{array}{c}{U}_F=80\text{N}\left(2\text{m}\right)\\ =160\text{N}\cdot \text{m}\times \frac{1\text{J}}{1\text{N}\cdot \text{m}}\\ \text{=}-\text{160}\text{J}\end{array}$

Thus, the work done by the force on the block is $-\text{160}\text{J}$.