# A 10-kg wagon moves horizontally at an initial speed of 5 m/s. A 30-N force is applied to the...

## Question:

A 10-kg wagon moves horizontally at an initial speed of 5 m/s. A 30-N force is applied to the wagon by pulling the rigid handle, which is angled 60{eq}^{\circ} {/eq} above the horizontal. The wagon continues to move horizontally for another 20 m. A negligible amount of work is converted into thermal energy.

By how much has the wagon's kinetic energy increased over the 20 m?

## Kinetic energy:

The kinetic energy of the object depends upon the mass of the object and the velocity of the object. The kinetic energy is half of the product of the mass of the object to the square of the velocity of the object.

## Answer and Explanation:

**Given Data**

- The force is {eq}F = 30\;{\rm{N}} {/eq}.

- The mass of the weapon is {eq}m = 10\;{\rm{kg}} {/eq}.

- The angle is {eq}\theta = 60^\circ {/eq}.

- The distance is {eq}d = 20\;{\rm{m}} {/eq}.

- The initial speed is {eq}u = 5\;{\rm{m/s}} {/eq}.

The expression for the force is,

{eq}F = ma {/eq}

Substitute the given values in the above expression.

{eq}\begin{align*} 30\cos 60^\circ &= 10\;{\rm{kg}} \times a\\ a &= \dfrac{{15}}{{10}}\\ &= 1.5\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{align*} {/eq}

The expression for the final velocity is,

{eq}{v^2} = {u^2} + 2ad {/eq}

Substitute the given values in the above expression.

{eq}\begin{align*} {v^2} &= {\left( {5\;{\rm{m/s}}} \right)^2} + 2\left( {1.5\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\left( {20\;{\rm{m}}} \right)\\ {v^2} &= 25\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}} + 60\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}\\ v &= \sqrt {85\;{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}} \\ &= 9.219\;{\rm{m/s}}\\ &\approx 9.2\;{\rm{m/s}} \end{align*} {/eq}

The expression for the change in kinetic energy is,

{eq}k = \dfrac{1}{2}m\left( {{v^2} - {u^2}} \right) {/eq}

Substitute the given values.

{eq}\begin{align*} k &= \dfrac{1}{2}\left( {10\;{\rm{kg}}} \right)\left( {{{\left( {9.219\;{\rm{m/s}}} \right)}^2} - {{\left( {5\;{\rm{m/s}}} \right)}^2}} \right)\\ &= 300\;{\rm{kg}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}\left( {\dfrac{{1\;{\rm{J}}}}{{1\;{\rm{kg}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{s}}^{\rm{2}}}}}} \right)\\ &= 300\;{\rm{J}} \end{align*} {/eq}

Thus, the change in kinetic energy is {eq}300\;{\rm{J}} {/eq}.

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from Physics for Kids

Chapter 1 / Lesson 9