Question

A \(47.2-\mathrm{kg}\) block of ice slides down an incline \(1.62 \mathrm{~m}\) long and \(0.902 \mathrm{~m}\) high. A worker pushes up on the ice parallel to the incline so that it slides down at constant speed. The coefficient of kinetic friction between the ice and the incline is \(0.110 .\) Find \((a)\) the force exerted by the worker, \((b)\) the work done by the worker on the block of ice, and \((c)\) the work done by gravity on the ice.

Short answer

The force exerted by the worker and the work done by gravity will be positive values since they are in the direction of displacement, while the work done by the worker will have a negative sign as it is against the direction of displacement. Exact values will be obtained after substituting the given data into the derived expressions.

Step-by-step solution

Calculate the angle of inclination

Considering the right triangle formed by the incline, the angle of inclination \( \theta \) can be calculated by tan inverse of the ratio of height to length. Hence, using the formula \( \theta = \tan^{-1}(\frac{0.902}{1.62}) \)

Calculate the force exerted by the worker

The block of ice slides down at constant speed. This means that the net force along the incline is zero. Therefore, the force exerted by the worker must balance the gravitational force and the frictional force. The force exerted by the worker is given by \( F_\text{worker} = F_\text{gravity} + F_\text{friction} = m \cdot g \cdot \sin\theta + \mu \cdot m \cdot g \cdot \cos\theta \) where m is the mass of the block, g is the acceleration due to gravity, \( \mu \) is the coefficient of friction, and \( \theta \) is the angle of the incline.

Calculate the work done by the worker and gravity

The work done by a force during displacement is given by \( W = F \cdot s \cdot \cos\alpha \) where F is the magnitude of the force, s is the displacement and \( \alpha \) is the angle between the force and the displacement. So, work done by the worker is \( W_\text{worker} = F_\text{worker} \cdot s \cdot \cos180^o = - F_\text{worker} \cdot s \) and the work done by gravity is \( W_\text{gravity}= m \cdot g \cdot h \)

Key concepts

Kinetic Friction

When an object moves across a surface, it experiences a force known as kinetic friction. This force opposes the motion, slowing the object down. The magnitude of kinetic friction is determined by the normal force acting on the object and the coefficient of kinetic friction (denoted as \( \mu_k \)). The formula to compute the kinetic friction is:

• \( F_\text{friction} = \mu_k \cdot F_\text{normal} \)

In inclined plane problems, like the one we are analyzing, the normal force is closely related to the weight of the object but acts perpendicular to the slope. The coefficient of kinetic friction, \( \mu_k \), is a dimensionless constant.
It varies depending on the types of surfaces in contact. In this exercise, the ice interacts with the incline, and the worker's push balances the forces so the ice moves at constant speed.

Constant Speed

When an object moves at constant speed, its velocity remains unchanged over time. This indicates that the acceleration is zero.
In our inclined plane problem, the block of ice slides down without accelerating because any increase in speed is countered by the worker.

• The forces acting on the object are balanced.
• Net force becomes zero when moving at a constant speed.

Gravity attempts to accelerate the block down the slope, but the worker's push and friction balance out this effect. As a result, the block doesn't speed up or slow down.
Understanding how forces cancel each other out here is crucial for grasping problems involving constant speed.

Work-Energy Principle

The work-energy principle states that work done by forces on an object results in a change in its kinetic energy. For the block of ice on the incline, several forces perform work: the worker, friction, and gravity. This principle helps describe how energy exchanges occur during motion.

• Work is calculated as \( W = F \cdot d \cdot \cos\alpha \).
• Where \( F \) is the force applied, \( d \) is the displacement, and \( \alpha \) is the angle between force and motion.

In this problem, since the block moves at constant speed, the net work done on it is zero, meaning the kinetic energy remains constant. The work by the worker is negative because it's against the direction of motion. Meanwhile, the work done by gravity aids the block's movement down the slope. The positive work done by gravity is balanced by the negative work from friction and the worker's force, resulting in no net energy change.

Force Balance

Forces acting on an object can be balanced or unbalanced. To move at a constant speed, as in our ice block example, the forces must be balanced. The sum of all forces acting on the block parallel to the incline equals zero.

• Gravitational force component along the slope: \( m \cdot g \cdot \sin\theta \)
• Kinetic friction: \( \mu \cdot m \cdot g \cdot \cos\theta \)
• Worker's force: \( F_\text{worker} \)

For balance,\[ F_\text{worker} + \mu \cdot m \cdot g \cdot \cos\theta = m \cdot g \cdot \sin\theta \]Each force plays a crucial role in maintaining constant speed. Gravitational force pulls the block down, friction resists motion, and the worker's force adjusts to exactly counter gravity and friction. This delicate balance ensures the block doesn't accelerate.