Chapter 14, Problem 1FP

To determine

The velocity of the block when $s=0.5\text{m}$.

Answer to Problem 1FP

The velocity of the block at the position of $s=0.5\text{m}$ is $5.24\text{m/s}$.

Explanation of Solution

Given:

The mass of the block is $m=1\text{0}\text{kg}$.

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

The stiffness of the spring is $k=500\text{N/m}$.

Initially the block is at rest $s_0=0$.

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

Write the formula to calculate the work done by the force $\left(F\right)$.

$U_{F\left(1-2\right)}=Fs$ (I)

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

Write the formula to calculate the work done by the spring force $\left(F_s\right)$.

$U_{F_s\left(1-2\right)}=-\frac{1}{2}k{s}^2$ (II)

Here, $k$ is the stiffness of the spring and $s$ is the displacement.

Write the formula for kinetic energy $\left(T\right)$.

$T=\frac{1}{2}m{v}^2$ (III)

Here, $m$ is the mass of the object and $v$ is the velocity of the object.

Write the formula for Principle of Work and Energy.

$T_1+{\displaystyle \sum U_{1-2}=T_2}$ (IV)

Here, $T_1$ is initial kinetic energy, $T_2$ is final kinetic energy and ${\displaystyle \sum U_{1-2}}$ is total work done by all the forces on the block.

Conclusion:

Calculate the Work done on the block while having the displacement of $0.5\text{m}$ by the

force $\left(F\right)$.

The force $\left(F\right)$ acting on the block at the slope of $\left(\frac{4}{5}\right)$.

Substitute $\left(500\times \frac{4}{5}\right)\text{N}$ for $F$ and $0.5\text{m}$ for $s$ in Equation (I).

$\begin{array}{c}{U}_{F\left(1-2\right)}=\left(500\times \frac{4}{5}\right)\text{N}\times 0.5\text{m}\\ \text{= 200}\text{J}\end{array}$

Calculate the Work done on the block while having the displacement of $0.5\text{m}$ by the

spring force $\left(F_s\right)$.

Substitute $500\text{N/m}$ for $k$ and $0.5\text{m}$ for $s$ in Equation (II).

$\begin{array}{c}{U}_{F_s\left(1-2\right)}=-\frac{1}{2}\left(500\text{N/m}\right){\left(0.5\text{m}\right)}^2\\ =-62.5\text{J}\end{array}$

Calculate total work done on the block $\left({\displaystyle \sum U_{1-2}}\right)$.

${\displaystyle \sum U_{1-2}}=U_{F\left(1-2\right)}+U_{F_s\left(1-2\right)}$

Substitute $200\text{J}$ for $U_{F\left(1-2\right)}$ and $-62.5\text{J}$ for $U_{F_s\left(1-2\right)}$.

$\begin{array}{c}{\displaystyle \sum U_{1-2}}=200\text{J + }\left(62.5\text{J}\right)\\ =200-125\\ =137.5\text{J}\end{array}$

Calculate the initial and final kinetic energy of the block.

At initial:

The block is at rest.

$T_1=0$

At final:

Substitute $T_2$ for $T$, $10\text{kg}$ for $m$ and $v_2$ for $v$ in Equation (III).

$\begin{array}{c}{T}_2=\frac{1}{2}\left(10\text{kg}\right)v_2{}^2\\ =5v_2{}^2\text{J}\end{array}$

Calculate the final velocity of the block at $s=0.5\text{m}$.

Substitute $0$ for $T_1$, $137.5\text{J}$ for ${\displaystyle \sum U_{1-2}}$ and $5v_2{}^2$ for $T_2$ in Equation (IV).

$\begin{array}{l}0+75\text{J}=5v_2{}^2\\ v_2{}^2=\frac{137.5}{5}\\ v_2=\sqrt{27.5}\\ =5.244\text{m/s}\end{array}$

Thus, the velocity of the block at the position of $s=0.5\text{m}$ is $5.24\text{m/s}$.