Newton's second law of motion:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$

Kinetic energy:

$\overline{){\mathbf{K}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}}^{{\mathbf{2}}}}$, where m is the mass and v is the velocity.

**Part A**

From Newton's second law, we can get acceleration, a, as:

$\overline{)\begin{array}{rcl}{\mathbf{F}}& {\mathbf{=}}& \mathbf{m}\mathbf{a}\\ {\mathbf{a}}& {\mathbf{=}}& \frac{\mathbf{F}}{\mathbf{m}}\end{array}}$

But we know that acceleration is the derivative of velocity with respect to time.

Thus,

$\begin{array}{rcl}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}& \mathbf{=}& \frac{\mathbf{F}}{\mathbf{m}}\\ \mathbf{(}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{s}}\mathbf{\times}\frac{\mathbf{d}\mathbf{s}}{\mathbf{d}\mathbf{t}}\mathbf{)}& \mathbf{=}& \frac{\mathbf{F}}{\mathbf{m}}\end{array}$

Differentiating displacement, s, with respect to time gives velocity.

$\begin{array}{rcl}\mathbf{v}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{s}}& \mathbf{=}& \frac{\mathbf{F}}{\mathbf{m}}\\ \mathbf{v}\mathbf{d}\mathbf{v}& \mathbf{=}& \frac{\mathbf{F}}{\mathbf{m}}\mathbf{d}\mathbf{s}\\ {\mathbf{\int}}_{{\mathbf{v}}_{\mathbf{i}}}^{{\mathbf{v}}_{\mathbf{f}}}\mathbf{v}\mathbf{d}\mathbf{v}& \mathbf{=}& \frac{\mathbf{F}}{\mathbf{m}}{\mathbf{\int}}_{\mathbf{0}}^{\mathbf{s}}\mathbf{d}\mathbf{s}\\ {\mathbf{\left[}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}}\mathbf{\right]}}_{{\mathbf{v}}_{\mathbf{i}}}^{{\mathbf{v}}_{\mathbf{f}}}& \mathbf{=}& \frac{\mathbf{F}}{\mathbf{m}}{\mathbf{\left[}\mathbf{s}\mathbf{\right]}}_{\mathbf{0}}^{\mathbf{s}}\\ \frac{{{\mathbf{v}}_{\mathbf{f}}}^{\mathbf{2}}\mathbf{-}{{\mathbf{v}}_{\mathbf{i}}}^{\mathbf{2}}}{\mathbf{2}}& \mathbf{=}& \frac{\mathbf{F}\mathbf{D}}{\mathbf{m}}\end{array}$

where s is the displacement D.

A particle of mass *M* moves along a straight line with initial speed *v*_{i}. A force of magnitude *F* pushes the particle a distance *D *along the direction of its motion.

Part A

Find v_{f}, the final speed of the particle after it has traveled a distance D. Express the final speed in terms of v_{i, }M, F and D.

Increase in mass

For the next two parts, assume that the particle's mass is increased to 3M, while all other parameters in the problem introduction remain the same.

Part B

By what multiplicative factor R_{k} does the initial kinetic energy increase, and by what multiplicative factor R_{w} does the work done by the force increase (with respect to the case when the particle had a mass M)?

If one of the quantities doubles, for instance, it would increase by a factor of 2. If a quantity stays the same, then the multiplicative factor would be 1.

Part C

The particle's change in speed over the distance D will be ______ the change in speed when it had a mass equal to M.

Increase in initial velocity

For the final two parts, assume that the initial speed of the particle is increased to 3_{vi}, with the particle's mass once again equal to M

part D

By what factor R_{k} does the initial kinetic energy increase (with respect to the first situation, with mass M and speed v_{i}), and by what factor Rw does the work done by the force increase?

part E

The particle's change in speed over the distance D will be ______ the change in speed when it had an initial velocity equal to v_{i}.

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