Question

The figure(Figure 1) shows the velocity graph of a 2.7kg object as it moves along the x-axis.

The figure(Figure 1) shows the velocity graph of a 2.7kg object as it moves along the x - axis. A) What is the net force acting on this object at t= 1 s? B) At 4 s? C) At 7 s?

A) What is the net force acting on this object at t= 1 s?

B) At 4 s?

C) At 7 s?

Answer 1

Concepts and reason

The concepts used to solve the problem are force and change in velocity with time.

Initially, find the slope of the increasing curve to find the acceleration of the object. Then, use the expression of force to find force acting on the object at 1 s.

Then, find the slope of the horizontal curve to find the acceleration of the object. Then, use the expression of force to find force acting on the object at 4 s.

Then, find the slope of the decreasing curve to find the acceleration of the object. Then, use the expression of force to find force acting on the object at 7 s.

Fundamentals

The acceleration of the object can be calculated by using the following expression:

$a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$

Here, ${v_1}$ is the velocity of the object at time ${t_1}$ and ${v_2}$ is the velocity of the object at time ${t_2}$ .

The expression of force is as follows:

$F = ma$

Here, F is the force. m is the mass, and a is the acceleration of the object.

(A)

The increasing curve starts from 0 s and end at 3 s. This means that the time 1 s lies between this interval.

Let $3{\rm{ s}}$ be the second point of the curve with velocity 12 m/s and 0 s be the first point of the curve with velocity 0 m/s.

The acceleration of the object in this interval can be calculated as follows:

$a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$

Here, ${v_1}$ is the first point of the curve at time ${t_1}$ and ${v_2}$ is the final point of the curve at time ${t_2}$ .

Substitute 12 m/s for ${v_2}$ , 0 m/s for ${v_1}$ , 3 s for ${t_2}$ , and 0 s for ${t_1}$ .

$\begin{array}{c}\\a = \frac{{12{\rm{ m/s}} - 0{\rm{ m/s}}}}{{3{\rm{ s}} - 0{\rm{ s}}}}\\\\ = 4{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The expression of force is as follows:

$F = ma$

Substitute 2.7 kg for m and $4{\rm{ m/}}{{\rm{s}}^2}$ for a in the above expression.

$\begin{array}{c}\\F = \left( {2.7{\rm{ kg}}} \right)\left( {4{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 10.8{\rm{ N}}\\\end{array}$

(B)

The horizontal curve starts from 3 s and end at 6 s. This means that the time 4 s lies between this interval.

Let ${\rm{6 s}}$ be the second point of the curve with velocity 12 m/s and 3 s be the first point of the curve with velocity 12 m/s.

The acceleration of the object in this interval can be calculated as follows:

$a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$

Here, ${v_1}$ is the first point of the curve at time ${t_1}$ and ${v_2}$ is the final point of the curve at time ${t_2}$ .

Substitute 12 m/s for ${v_2}$ , 12 m/s for ${v_1}$ , 6 s for ${t_2}$ , and 3 s for ${t_1}$ .

$\begin{array}{c}\\a = \frac{{12{\rm{ m/s}} - 12{\rm{ m/s}}}}{{{\rm{6 s}} - 3{\rm{ s}}}}\\\\ = 0{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The expression of force is as follows:

$F = ma$

Substitute 2.7 kg for m and ${\rm{0 m/}}{{\rm{s}}^2}$ for a in the above expression.

$\begin{array}{c}\\F = \left( {2.7{\rm{ kg}}} \right)\left( {{\rm{0 m/}}{{\rm{s}}^2}} \right)\\\\ = 0.0{\rm{ N}}\\\end{array}$

(C)

The decreasing curve starts from 6 s and end at 8 s. This means that the time 7 s lies between this interval.

Let 8 s be the second point of the curve with velocity 0 m/s and 6 s be the first point of the curve with velocity 12 m/s.

The acceleration of the object in this interval can be calculated as follows:

$a = \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}$

Here, ${v_1}$ is the first point of the curve at time ${t_1}$ and ${v_2}$ is the final point of the curve at time ${t_2}$ .

Substitute 0 m/s for ${v_2}$ , 12 m/s for ${v_1}$ , 8 s for ${t_2}$ , and 6 s for ${t_1}$ .

$\begin{array}{c}\\a = \frac{{{\rm{0 m/s}} - 12{\rm{ m/s}}}}{{{\rm{8 s}} - 6{\rm{ s}}}}\\\\ = - 6{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

The expression of force is as follows:

$F = ma$

Substitute 2.7 kg for m and $- {\rm{6 m/}}{{\rm{s}}^2}$ for a in the above expression.

$\begin{array}{c}\\F = \left( {2.7{\rm{ kg}}} \right)\left( {{\rm{ - 6 m/}}{{\rm{s}}^2}} \right)\\\\ = - 16.2{\rm{ N}}\\\end{array}$

Ans: Part A

The net force acting on the object at 1 s is 10.8 N.

Part B

The net force acting on the object at 4 s is 0.0 N.

Part C

The net force acting on the object at 7 s is -16.2 N.