91Ó°ÊÓ

The gravitational force is the force exerted by the Earth that pulls objects towards its center. In the context of inclined planes, it's crucial to understand how this force affects an object. The gravitational force acting on the block of ice is calculated using the component of the force parallel to the incline. This is given by the equation: \( F_g = mg \sin \theta \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline. For the given exercise:
- Mass, \( m = 45 \mathrm{~kg}\)
- The acceleration due to gravity, \( g = 9.8 \mathrm{~m/s}^2\)
- \(\theta = \arcsin \left(\frac{0.91}{1.5}\right) \approx 38\degree \) The gravitational force pulling the block down the incline is:
\( F_g = 45 \mathrm{~kg} \cdot 9.8 \mathrm{~m/s}^2 \cdot \frac{0.91}{1.5} = 269.94 \mathrm{~N} \. \)

The normal force is the support force exerted perpendicularly by a surface, which prevents an object from penetrating it. On an inclined plane, it's important to understand that the normal force acts perpendicular to the surface of the incline and not directly opposite to the gravitational force. This force can be calculated by the following equation: \( F_n = mg \cos \theta \), where \( m \) is the mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline. For our given problem:
- \( m = 45 \mathrm{~kg} \)
- \( g = 9.8 \mathrm{~m/s}^2 \)
- \(\theta = \arcsin \left(\frac{0.91}{1.5} \right) \approx 38 \degree \) The normal force is given by: \( F_n = 45 \mathrm{~kg} \cdot 9.8 \mathrm{~m/s}^2 \cdot \cos 38 \degree = 355.1 \mathrm{~N} \). The crucial point here is that since the normal force is perpendicular to the motion (which is parallel to the incline), it does no work on the block. Therefore, the work done by the normal force \( W_n = 0 \).

Work is done when a force is applied to an object and it moves. The amount of work done is calculated by the equation: \( W = F \cdot d \cdot \cos \theta \. \) For the worker's force:
- Force, \( F = 269.94 \mathrm{~N} \)
- Distance, \( d = 1.5 \mathrm{~m} \)
- Angle, \( \theta = 0 \degree \) Substituting these values, we get: \( W = 269.94 \mathrm{~N} \cdot 1.5 \mathrm{~m} \cdot \cos 0 \degree = 404.91 \mathrm{~J} \). For the gravitational force:
- Force, \( F_g = 45 \mathrm{~kg} \cdot 9.8 \mathrm{~m/s}^2 \cdot \frac{0.91}{1.5} = 269.94 \mathrm{~N}\)
- Height, \( h = 0.91 \mathrm{~m} \)
The work done by the gravitational force is: \( W_g = mgh = 45 \mathrm{~kg} \cdot 9.8 \mathrm{~m/s}^2 \cdot 0.91 \mathrm{~m} = 400.95 \mathrm{~J} \). With the block moving at a constant speed, the net work done on it is zero: \( W_{net} = 0 \). Therefore, we conclude that:

• Worker’s force work: 404.91 J
• Gravitational work: 400.95 J
• Normal force work: 0 J.

Understanding these components helps you grasp how forces interact on an inclined plane and the work done by each force.