Chapter 4, Problem 1SP

(a)

To determine

The horizontal acceleration of the block if a horizontal force of $30\text{N}$ is exerted by a string attached to a block being pulled across a tabletop and it also experiences a frictional force of $5\text{N}$ due to contact with the table.

Answer to Problem 1SP

The horizontal acceleration of the block if a horizontal force of $30\text{N}$ is exerted by a string attached to a $5\text{kg}$ block being pulled across a tabletop is $5\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The horizontal force is $30\text{N}$, the mass of the block is $5\text{kg}$ and frictional force is $5\text{N}$.

Write the expression for the net horizontal force.

$F_{\text{net}}=F_{\text{tension}}-F_{\text{friction}}$

Here,

$F_{\text{net }}$ is the net force acting on the block

$F_{\text{tension}}$ is the horizontal force

$F_{\text{friction}}$ is the frictional force

The negative sign indicate that frictional force is opposite to horizontal force

Substitute $30\text{N}$ for $F_{\text{tension}}$ and $5\text{N}$ for $F_{\text{friction}}$ in the above equation to get $F_{\text{net}}$.

$\begin{array}{c}{F}_{\text{net}}=30\text{N}-5\text{N}\\ =25\text{N}\end{array}$

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=\frac{F_{\text{net}}}{m}$

Here,

$m$ is the mass of the block

$a$ is the horizontal acceleration of the block

Substitute $22\text{N}$ for $F_{\text{net}}$ and $8\text{kg}$ for $m$ in the above equation to get $a$.

$\begin{array}{l}a=\frac{25\text{N}}{5\text{kg}}\\ =5.0\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the horizontal acceleration of the block if a horizontal force of $30\text{N}$ is exerted by a string attached to a $5\text{kg}$ block being pulled across a tabletop is $5\text{.0}\text{m}/{\text{s}}^{\text{2}}$.

(b)

To determine

Velocity of the block after $3$ seconds from rest.

Answer to Problem 1SP

Velocity of the block after $3$ seconds from rest is $15\text{m}/\text{s}$.

Explanation of Solution

Given info: The time after which velocity is to be find is $3\text{s}$.

Write the expression for the equation of motion of the block.

$v=v_0+at$

Here,

$v$ is the velocity of the block after $t$ second

$v_0$ is the initial velocity

$a$ is the acceleration

Substitute $0\text{m}/{\text{s}}^{\text{2}}$ for $v_0$ ,$3\text{s}$ for $t$ and $5.0\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to get $v$.

$\begin{array}{c}v=0\text{m}/{\text{s}}^{\text{2}}+\left(5.0\text{m}/{\text{s}}^{\text{2}}\right)\left(3\text{s}\right)\\ =15\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of the block after $3$ seconds from rest is $15\text{m}/\text{s}$.

(c)

To determine

The distance it travels in $3$ seconds.

Answer to Problem 1SP

The distance it travels in $3$ seconds is $22.5\text{m}$.

Explanation of Solution

Write the expression for the distance travelled by the block.

$d=v_{0}t+\frac{1}{2}a{t}^2$

Here,

$d$ is the distance

$a$ is the acceleration

$t$ is the time

Substitute $0\text{m}/\text{s}$ for $v_0$ ,$5.0\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $3\text{s}$ for $t$ in the above equation to get $s$.

$\begin{array}{c}d=\left(0\text{m}/\text{s}\right)\left(3\text{s}\right)+\frac{1}{2}\left(5.0\text{m}/{\text{s}}^{\text{2}}\right){\left(3\text{s}\right)}^2\\ =22.5\text{m}\end{array}$

Conclusion:

Thus, the distance it travels in $3$ seconds is $22.5\text{m}$.