Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

To determine

(a)

The velocity of the block when $t=2\text{s}$.

Answer to Problem 1PP

The velocity of the block at the time $t=2\text{s}$ is $20\text{m/s}$.

Explanation of Solution

Given:

The mass of the block is $m=10\text{kg}$.

Initially the velocity of the block is $v=0$ at $t=0$.

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

Write the formula for Newton’s second law of motion along $x-axis$.

${\displaystyle \sum F_x=m{a}_x}$ (I)

Here, $m$ is the mass of the block, $a_x$ is the acceleration of the block along $x-axis$ and ${\displaystyle \sum F_x}$ is the resultant of all the forces acting on the block along $x-axis$.

Write the formula for velocity of the block at constant acceleration as a function of time.

$v=v_0+a_{c}t$ (II)

Here, $v$ is the final velocity, $v_0$ is the initial velocity, $a_c$ is the constant acceleration and $t$ is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along $x-axis$.

$\begin{array}{c}{\displaystyle \sum F_x}=500\text{N}\left(\frac{4}{5}\right)-300\text{N}\\ \text{=100}\text{N}\end{array}$

Calculate the acceleration of the block.

Substitute $100\text{N}$ for ${\displaystyle \sum F_x}$, $10\text{kg}$ for $m$ in Equation (I).

$\begin{array}{l}100\text{N}=\left(10\text{kg}\right)a_x\\ a_x=\frac{100}{10}\\ a_x=10{\text{m/s}}^2\end{array}$

Calculate the velocity of the block at the time $t=2\text{s}$.

Here, $a_x=a_c=10{\text{m/s}}^2$.

Substitute $0$ for $v_0$, $10{\text{m/s}}^2$ for $a_c$ and $2\text{s}$ for $t$ in Equation (II).

$\begin{array}{c}v=0+\left(10{\text{m/s}}^2\right)\left(2\text{s}\right)\\ =0+20\\ =20\text{m/s}\end{array}$

Thus, the velocity of the block at the time $t=2\text{s}$ is $20\text{m/s}$.

To determine

b)

The velocity of the block when $t=2\text{s}$.

Answer to Problem 1PP

The velocity of the block at the time $t=2\text{s}$ is $4\text{m/s}$.

Explanation of Solution

Given:

The mass of the block is $m=10\text{kg}$.

Initially the velocity of the block is $v=0$ at $t=0$.

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

Write the formula for Newton’s second law of motion along $x-axis$.

${\displaystyle \sum F_x=m{a}_x}$ (I)

Here, $m$ is the mass of the block, $a_x$ is the acceleration of the block along $x-axis$ and ${\displaystyle \sum F_x}$ is the resultant of all the forces acting on the block along $x-axis$.

Write the formula for acceleration $\left(a\right)$ of the block.

$\begin{array}{l}a=\frac{dv}{dt}\\ dv=adt\end{array}$ (II)

Here, $\frac{dv}{dt}$ is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along $x-axis$.

$\begin{array}{c}{\displaystyle \sum F_x}=20t\\ \text{= 20}t\end{array}$

Calculate the acceleration of the block.

Substitute $20t$ for ${\displaystyle \sum F_x}$, $10\text{kg}$ for $m$ in Equation (I).

$\begin{array}{l}20t=\left(10\text{kg}\right)a_x\\ a_x=\frac{20t}{10}\\ a_x=2t\end{array}$

Calculate the velocity of the block at the time $t=2\text{s}$.

Here, $a_x=a=2t$.

Substitute $2t$ for $a$ in Equation (II).

$dv=2tdt$

Integrate at the limits of $v=0\text{to }v$ and $t=0\text{to}\text{2}\text{s}$.

$\begin{array}{l}{\displaystyle \underset{0}{\overset{v}{\int }}dv}={\displaystyle \underset{0}{\overset{2}{\int }}2tdt}\\ {\left[v\right]}_0^v={\left[\frac{2t^2}{2}\right]}_0^2\\ \left(v-0\right)={\left(2\right)}^2-{\left(0\right)}^0\\ v=4\text{m/s}\end{array}$

Thus, the velocity of the block at the time $t=2\text{s}$ is $4\text{m/s}$.