The work done by the tension force in the rope, pulling a 100-kg object vertically upward by 5.0 m with an acceleration of 1.0 m/s², is 500 J.
To find the work done by the tension force in the rope, we can use the formula for work:
[tex]\[ W = F \times d \times \cos(\theta) \][/tex]
Where:
- [tex]\( W \)[/tex] is the work done,
- [tex]\( F \)[/tex] is the force applied (in the direction of the displacement),
- [tex]\( d \)[/tex] is the displacement, and
- [tex]\( \theta \)[/tex] is the angle between the force and the direction of displacement.
In this scenario:
- [tex]\( F \)[/tex] is the tension force in the rope,
- [tex]\( d \)[/tex] is the displacement of the object, and
- [tex]\( \theta \)[/tex] is the angle between the tension force and the direction of displacement, which is 0° since the force and displacement are in the same direction.
Given:
- Mass of the object, [tex]\( m = 100 \, \text{kg} \)[/tex]
- Acceleration, [tex]\( a = 1.0 \, \text{m/s}^2 \)[/tex]
- Displacement, [tex]\( d = 5.0 \, \text{m} \)[/tex]
First, let's find the tension force using Newton's second law:
[tex]\[ F = m \times a \][/tex]
[tex]\[ F = (100 \, \text{kg}) \times (1.0 \, \text{m/s}^2) \][/tex]
[tex]\[ F = 100 \, \text{N} \][/tex]
Now, we can calculate the work done by the tension force:
[tex]\[ W = F \times d \times \cos(\theta) \][/tex]
[tex]\[ W = (100 \, \text{N}) \times (5.0 \, \text{m}) \times \cos(0^{\circ}) \][/tex]
[tex]\[ W = (100 \, \text{N}) \times (5.0 \, \text{m}) \times 1 \][/tex]
[tex]\[ W = 500 \, \text{J} \][/tex]
So, the work done by the tension force in the rope is [tex]\( 500 \, \text{J} \).[/tex]
