91Ó°ÊÓ

In physics, the tension force is the force that is transmitted through a string, cable, or wire when it's pulled tight by forces acting from opposite ends. In the given exercise, the tension force is the force exerted by the cable of the tow truck to pull the car. It's important to note that tension is always directed along the length of the cable and is constant throughout if the cable is considered massless and inextensible.
To calculate the work done by the tension force, we utilize the formula for work, which is the dot product of force and displacement vectors:

• Work, \( W = F \, \cdot \, d \cdot \cos(\theta) \)
• \( F \) is the magnitude of the tension force
• \( d \) is the displacement of the car
• \( \theta \) is the angle between the force direction and displacement direction

In scenarios where tension is applied horizontally, the angle \( \theta \) is zero. This simplification leads to maximum work being done by the tension force.

A horizontal pull refers to the force applied parallel to the surface of the ground. In our context, it is when the car is pulled directly along the horizontal roadway by the tow truck.
When analyzing work done through horizontal pull, the absence of vertical displacement implies that gravitational forces don't need to be considered in this direction. The formulas used become:

• \( W = F \, \cdot \, d \)
• Since \( \cos(0^{\circ}) = 1 \), the equation simplifies to \( W = F \, \cdot \, d \)

This approach makes calculations straightforward, as the entire tension force contributes to moving the car. This is different from when other forces, like those acting at an angle from the surface, are involved.

The angle of force application is crucial in determining how much of a force actually contributes to doing work. When a force is applied at an angle, only the component of the force that acts in the direction of the motion does work.
To break this down:

• When the force is at an angle \( \theta \), only \( F \cdot \cos(\theta) \) of the force contributes towards work.
• The work done \( W \) is given as \( W = F \cdot d \cdot \cos(\theta) \)

In the original problem, when the force is at \( 35^{\circ} \) above the horizontal, only the horizontal component of the tension does work. This reduction in effective force reduces the overall work done, emphasizing how angle plays a significant role in such scenarios.

Gravitational work refers to the work done by the force of gravity, which is essentially the force acting downwards. This work is calculated when there is a vertical component of movement against or with the gravitational pull.
In the given problem, since the car is moved horizontally, there is no vertical displacement, meaning:

• The work done by gravity would be \( W_{gravity} = 0 \)
• This is because gravitational work requires vertical movement, which does not occur along a flat plane

Under these conditions, while gravity may still be acting on the car, it does not do any work on the car, simplifying the energy considerations in this horizontal movement case.