Question
Question 5
Use the formula above to calculate work done.
Calculate the amount of work done by the force in moving the object through a distance of 9 m if an object is horizontally dragged across the surface by a 130 N force acting parallel to the surface.
a
507 J
b
.070 J
c
1170 J
d
14 J
Use the formula above to calculate work done.
Calculate the amount of work done by the force in moving the object through a distance of 9 m if an object is horizontally dragged across the surface by a 130 N force acting parallel to the surface.
a
507 J
b
.070 J
c
1170 J
d
14 J
Answers
GPT-4o mini
To calculate the work done by a force acting on an object, you can use the formula:
\[ \text{Work} (W) = \text{Force} (F) \times \text{Distance} (d) \times \cos(\theta) \]
Where:
- \( W \) is the work done (in joules, J)
- \( F \) is the force applied (in newtons, N)
- \( d \) is the distance moved by the object (in meters, m)
- \( \theta \) is the angle between the force and the direction of motion.
In this case:
- The force \( F = 130 \, \text{N} \)
- The distance \( d = 9 \, \text{m} \)
- The force is acting parallel to the surface, so \( \theta = 0^\circ \) and \( \cos(0) = 1 \).
Now, substitute the values into the formula:
\[
W = 130 \, \text{N} \times 9 \, \text{m} \times \cos(0)
\]
\[
W = 130 \, \text{N} \times 9 \, \text{m} \times 1
\]
\[
W = 130 \times 9 = 1170 \, \text{J}
\]
Therefore, the amount of work done is **1170 J**, corresponding to option **c**.
\[ \text{Work} (W) = \text{Force} (F) \times \text{Distance} (d) \times \cos(\theta) \]
Where:
- \( W \) is the work done (in joules, J)
- \( F \) is the force applied (in newtons, N)
- \( d \) is the distance moved by the object (in meters, m)
- \( \theta \) is the angle between the force and the direction of motion.
In this case:
- The force \( F = 130 \, \text{N} \)
- The distance \( d = 9 \, \text{m} \)
- The force is acting parallel to the surface, so \( \theta = 0^\circ \) and \( \cos(0) = 1 \).
Now, substitute the values into the formula:
\[
W = 130 \, \text{N} \times 9 \, \text{m} \times \cos(0)
\]
\[
W = 130 \, \text{N} \times 9 \, \text{m} \times 1
\]
\[
W = 130 \times 9 = 1170 \, \text{J}
\]
Therefore, the amount of work done is **1170 J**, corresponding to option **c**.