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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?

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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.

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

[katex]a = \frac{{{v_2} – {v_1}}}{{{t_2} – {t_1}}}[/katex]

Here, [katex]{v_1}[/katex]

The expression of force is as follows:

[katex]F = ma[/katex]

Here,

Fis the force.mis the mass, andais 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 [katex]3{\rm{ s}}[/katex]

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

[katex]a = \frac{{{v_2} – {v_1}}}{{{t_2} – {t_1}}}[/katex]

Here, [katex]{v_1}[/katex]

Substitute 12 m/s for [katex]{v_2}[/katex]

[katex]\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}[/katex]

The expression of force is as follows:

[katex]F = ma[/katex]

Substitute 2.7 kg for m and [katex]4{\rm{ m/}}{{\rm{s}}^2}[/katex]

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

(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 [katex]{\rm{6 s}}[/katex]

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

[katex]a = \frac{{{v_2} – {v_1}}}{{{t_2} – {t_1}}}[/katex]

Here, [katex]{v_1}[/katex]

Substitute 12 m/s for [katex]{v_2}[/katex]

[katex]\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}[/katex]

The expression of force is as follows:

[katex]F = ma[/katex]

Substitute 2.7 kg for m and [katex]{\rm{0 m/}}{{\rm{s}}^2}[/katex]

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

(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:

[katex]a = \frac{{{v_2} – {v_1}}}{{{t_2} – {t_1}}}[/katex]

Here, [katex]{v_1}[/katex]

Substitute 0 m/s for [katex]{v_2}[/katex]

[katex]\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}[/katex]

The expression of force is as follows:

[katex]F = ma[/katex]

Substitute 2.7 kg for m and [katex]- {\rm{6 m/}}{{\rm{s}}^2}[/katex]

[katex]\begin{array}{c}\\F = \left( {2.7{\rm{ kg}}} \right)\left( {{\rm{ – 6 m/}}{{\rm{s}}^2}} \right)\\\\ = – 16.2{\rm{ N}}\\\end{array}[/katex]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.