High School

How much work must be done on a 6000-kg car to increase its speed from 5 m/s to 10 m/s?

A. 75000 J
B. 300000 J
C. -60000 J
D. 225000 J

Answer :

To solve this problem, we need to determine how much work is required to increase the speed of a 6000-kg car from 5 m/s to 10 m/s. Work done on an object to change its speed can be calculated using the change in kinetic energy.

Here are the steps to find the solution:

1. Calculate the initial kinetic energy:
The kinetic energy (KE) of an object is given by the formula:
[tex]\[
\text{KE} = \frac{1}{2} m v^2
\][/tex]
where [tex]\( m \)[/tex] is the mass of the object and [tex]\( v \)[/tex] is its velocity.

For the initial speed:
[tex]\[
\text{KE}_{\text{initial}} = \frac{1}{2} \times 6000 \, \text{kg} \times (5 \, \text{m/s})^2
\][/tex]
[tex]\[
\text{KE}_{\text{initial}} = \frac{1}{2} \times 6000 \times 25
\][/tex]
[tex]\[
\text{KE}_{\text{initial}} = 0.5 \times 6000 \times 25 = 75000 \, \text{J}
\][/tex]

2. Calculate the final kinetic energy:
Using the same formula for the final speed:
[tex]\[
\text{KE}_{\text{final}} = \frac{1}{2} \times 6000 \, \text{kg} \times (10 \, \text{m/s})^2
\][/tex]
[tex]\[
\text{KE}_{\text{final}} = \frac{1}{2} \times 6000 \times 100
\][/tex]
[tex]\[
\text{KE}_{\text{final}} = 0.5 \times 6000 \times 100 = 300000 \, \text{J}
\][/tex]

3. Calculate the work done:
The work done on the car to increase its speed is the change in kinetic energy:
[tex]\[
\text{Work} = \text{KE}_{\text{final}} - \text{KE}_{\text{initial}}
\][/tex]
[tex]\[
\text{Work} = 300000 \, \text{J} - 75000 \, \text{J}
\][/tex]
[tex]\[
\text{Work} = 225000 \, \text{J}
\][/tex]

Therefore, the work done to increase the speed of the car from 5 m/s to 10 m/s is 225000 J.

The correct answer is:
D. 225000 J