Question
a 10 kg object experiences a horisontal force of 10N which cause it to accelerate at 5 m/s2, moving it a distance of 20 m, horizontally. How much work is done by the force? (please show your work)
Answers
GPT-4o mini
To calculate the work done by the force, we can use the formula for work:
\[
\text{Work} (W) = \text{Force} (F) \times \text{Distance} (d) \times \cos(\theta)
\]
where:
- \( W \) is the work done,
- \( F \) is the force applied,
- \( d \) is the distance over which the force is applied,
- \( \theta \) is the angle between the force and the direction of motion.
In this case, we have:
- \( F = 10 \, \text{N} \) (the horizontal force applied),
- \( d = 20 \, \text{m} \) (the distance over which the force is applied),
- Since the force is in the same direction as the motion, \( \theta = 0^\circ \).
Now, we can plug in the values:
\[
\cos(0^\circ) = 1
\]
\[
W = 10 \, \text{N} \times 20 \, \text{m} \times 1
\]
\[
W = 200 \, \text{J}
\]
Therefore, the work done by the force is **200 Joules**.
\[
\text{Work} (W) = \text{Force} (F) \times \text{Distance} (d) \times \cos(\theta)
\]
where:
- \( W \) is the work done,
- \( F \) is the force applied,
- \( d \) is the distance over which the force is applied,
- \( \theta \) is the angle between the force and the direction of motion.
In this case, we have:
- \( F = 10 \, \text{N} \) (the horizontal force applied),
- \( d = 20 \, \text{m} \) (the distance over which the force is applied),
- Since the force is in the same direction as the motion, \( \theta = 0^\circ \).
Now, we can plug in the values:
\[
\cos(0^\circ) = 1
\]
\[
W = 10 \, \text{N} \times 20 \, \text{m} \times 1
\]
\[
W = 200 \, \text{J}
\]
Therefore, the work done by the force is **200 Joules**.