Question

(a) What is the kinetic energy in joules of a 1060-kg automobile traveling at 94 km/h? Incorrect: Your answer is incorrect. answer in:___J(b) How much work would have to be done to bring a 1060-kg automobile traveling at 94 km/h to a stop? answer in:____J

Answer 1

Given:

The mass of the automobile is m = 1060 kg

The speed of the automobile is

$\begin{gathered} v=\text{ 94 km/h} \\ =\frac{94km}{h}\times\frac{1000\text{ m}}{1\text{ km}}\times\frac{1\text{ h}}{3600\text{ s}} \\ =26.11\text{ m/s} \end{gathered}$

Required:

(a)The kinetic energy

(b) Work done to bring the automobile to stop.

Explanation:

(a) The kinetic energy can be calculated by the formula

$KE\text{ = }\frac{1}{2}mv^2$

On substituting the values, the kinetic energy will be

$\begin{gathered} K.E.=\frac{1}{2}\times1060\times(26.11)^2\text{ } \\ =361318.013\text{ J} \end{gathered}$

(b) Work done is defined as the change in kinetic energy.

To stop the automobile, the kinetic energy should be zero.

So, the work done can be calculated as

$\begin{gathered} Work\text{ done = change in kinetic energy} \\ W=0-361318.013\text{ J} \\ W\text{ = 361318.013 J} \end{gathered}$

Final Answer:

(a)The kinetic energy is 361318.013 J

(b) Work done to bring the automobile to stop is 361318.013 J