# A little red wagon with mass 7.00 kg moves in a straight line on a frictionless horizontal...

## Question:

A little red wagon with mass {eq}7.00 kg{/eq} moves in a straight line on a frictionless horizontal surface. It has an initial speed of {eq}4.00 m/s{/eq} and then is pushed {eq}3.0 m{/eq} in the direction of the initial velocity by a force with a magnitude of {eq}10.0 N {/eq}

a. Use the work energy therorem to calculate the wagon's final speed.

b. Calculate the acceleration produced by the force. Use this acceleration in the kinematic relationships to calculate the wagon's final speed. Compare this result to that calculated in part a.

## The Work-Energy Theorem:

It states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

## Answer and Explanation:

**Part(a)**

Work done by the force is the product of force and distance i.e. 10(3) = 30J

This energy **W** is the change in kinetic energy **Ek** of the wagon and thus:

$$\begin{align*} \Delta E_{k} &= W\\ \frac{1}{2}m(v^2 - u^2) &= W\\ \frac{1}{2}(7)(v^2 - 4^2) &= 30\\ v^2 &= 30\cdot (\frac{2}{7}) +16\\ v &=\sqrt{24.57}\\ &= 4.96m/s \end{align*} $$

**Part(b)**

The acceleration due to the **10N** force is the ratio of the force to the mass i.e. {eq}a = \frac{10}{7} = 1.43m/s^2 {/eq}

- The initial velocity is
**4m/s** - The distance is
**3m**

$$\begin{align*} v^2 &= u^2 + 2as\\ &= \sqrt{u^2 + 2as}\\ &= \sqrt{4^2 + 2(1.43)(3)}\\ &= 4.96m/s \end{align*} $$

**As expected, the velocity is the same as from the Work-Energy condsideration**

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from CSET Science Subtest I - General Science (215): Practice & Study Guide

Chapter 10 / Lesson 6