Question

The vx-versus-time graph for a 1.8-kg object is shown in (Figure 1) . A single force acts on the object, and the force has only an x component.

Part A

What is Fx at t=0.50s?

Express your answer to two significant figures and include appropriate units.

Part B

What is Fx at t=2.0s?

Express your answer to two significant figures and include appropriate units.

Part C

What is Fx at t=4.0s?

Express your answer to two significant figures and include appropriate units.

Please show work, Thank you

Answer 1

Concepts and reason

The concept used to solve this problem are slope of the curve and Newton’s second law of motion.

Initially, calculate the force at $t = 0.5\;{\rm{s}}$ by using the concepts of slope of the curve and Newton’s second law of motion.

Later, calculate the force at $t = 2.0\;{\rm{s}}$ by using the concepts of slope of the curve and Newton’s second law of motion.

Finally, calculate the force at $t = 4.0\;{\rm{s}}$ by using the concepts of slope of the curve and Newton’s second law of motion.

Fundamentals

The momentum of the body is equal to the product of the mass of the body and its acceleration.

That is, the expression for the momentum of the body is as follows:

$p = mv$

Here, m is the mass of the body and v is the velocity of the body.

The expression for the net force acting on the system is as follows:

${F_{net}} = \frac{{dp}}{{dt}}$

Here, $dp$ is the change in momentum of the particle and $dt$ is the change in time.

In other words, the net force acting on the particle is as follows:

$\begin{array}{c}\\{F_{net}} = \frac{{d\left( {mv} \right)}}{{dt}}\\\\ = m\frac{{dv}}{{dt}}\\\end{array}$

(A)

From the graph, the slope of the curve at $t = 0.5\;{\rm{s}}$ is as follows:

$\frac{{dv}}{{dt}} = 4\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

The expression for the force acting on the body at $t = 0.5\;{\rm{s}}$ is as follows:

${F_{t = 0.5\;{\rm{s}}}} = m\frac{{dv}}{{dt}}$

Substitute 1.8 kg for m and $4\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for $\frac{{dv}}{{dt}}$ in the equation ${F_{t = 0.5\;{\rm{s}}}} = m\frac{{dv}}{{dt}}$ .

$\begin{array}{c}\\{F_{t = 0.5\;{\rm{s}}}} = \left( {1.8\;{\rm{kg}}} \right)\left( {4\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\\\ = 7.2\;{\rm{N}}\\\end{array}$

(B)

From the graph, the slope of the curve at $t = 2.0\;{\rm{s}}$ is as follows:

$\frac{{dv}}{{dt}} = - 1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$

The expression for the force acting on the body at $t = 2.0\;{\rm{s}}$ is as follows:

${F_{t = 2.0\;{\rm{s}}}} = m\frac{{dv}}{{dt}}$

Substitute 1.8 kg for m and $- 1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ for $\frac{{dv}}{{dt}}$ in the equation ${F_{t = 2.0\;{\rm{s}}}} = m\frac{{dv}}{{dt}}$ .

$\begin{array}{c}\\{F_{t = 2.0\;{\rm{s}}}} = \left( {1.8\;{\rm{kg}}} \right)\left( { - 1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\\\\ = - 1.8\;{\rm{N}}\\\end{array}$

(C)

From the graph, the slope of the curve at $t = 4.0\;{\rm{s}}$ is as follows:

$\frac{{dv}}{{dt}} = 0$

The expression for the force acting on the body at $t = 4.0\;{\rm{s}}$ is as follows:

${F_{t = 4.0\;{\rm{s}}}} = m\frac{{dv}}{{dt}}$

Substitute 1.8 kg for m and 0 for $\frac{{dv}}{{dt}}$ in the equation ${F_{t = 4.0\;{\rm{s}}}} = m\frac{{dv}}{{dt}}$ .

$\begin{array}{c}\\{F_{t = 4.0\;{\rm{s}}}} = \left( {1.8\;{\rm{kg}}} \right)\left( 0 \right)\\\\ = 0\\\end{array}$

Ans: Part A

The force acting on the body at $t = 0.5\;{\rm{s}}$ is $7.2\;{\rm{N}}$ .