A block of weight w slides down a rough inclined plane of angle θ. A constant frictional force f_k acts on the block so that it moves at a constant velocity v down the incline. Fin... A block of weight w slides down a rough inclined plane of angle θ. A constant frictional force f_k acts on the block so that it moves at a constant velocity v down the incline. Find the work done by each force acting on the block and the total work done on it as it descends a height h down the incline.
Understand the Problem
The question is asking to analyze the forces acting on a block sliding down an inclined plane, specifically focusing on the work done by frictional and gravitational forces as the block moves down a certain height while maintaining a constant velocity.
Answer
The net work done is given by: $$ W_\text{net} = mgh - \mu mg \cos(\theta) \left(\frac{h}{\sin(\theta)}\right) $$
Answer for screen readers
The expression for net work done, $W_\text{net}$, by gravitational and frictional forces is given by: $$ W_\text{net} = mgh - \mu mg \cos(\theta) \left(\frac{h}{\sin(\theta)}\right) $$
Steps to Solve
- Identify the forces acting on the block
The block on the inclined plane experiences two primary forces: gravitational force ($F_g$) and frictional force ($F_f$).
The gravitational force can be described as: $$ F_g = mg $$ where $m$ is the mass of the block and $g$ is the acceleration due to gravity.
The frictional force can be calculated using: $$ F_f = \mu N $$ where $\mu$ is the coefficient of friction and $N$ is the normal force.
- Calculate the normal force
Since the block is on an incline, the normal force is given by the component of gravitational force perpendicular to the inclined surface. It can be calculated using: $$ N = mg \cos(\theta) $$ where $\theta$ is the angle of the incline.
- Determine the work done by the gravitational force
The work done by the gravitational force as the block slides down can be found using the equation: $$ W_g = mgh $$ where $h$ is the vertical height the block descends.
- Determine the work done by the frictional force
Work done by friction is calculated as: $$ W_f = -F_f d $$ where $d$ is the distance traveled along the incline. The negative sign indicates that friction opposes the motion.
- Express $d$ in terms of $h$
Since the block descends a height $h$, the distance along the incline can be expressed as: $$ d = \frac{h}{\sin(\theta)} $$
- Combine the work done equations
Substituting the expressions for $W_g$ and $W_f$, we can summarize the work done as: $$ W_\text{net} = W_g + W_f = mgh - F_f d $$
Substituting the expression for $d$: $$ W_\text{net} = mgh - F_f \left( \frac{h}{\sin(\theta)} \right) $$
- Final expression
The net work can provide insight into the energy balance of the system.
The expression for net work done, $W_\text{net}$, by gravitational and frictional forces is given by: $$ W_\text{net} = mgh - \mu mg \cos(\theta) \left(\frac{h}{\sin(\theta)}\right) $$
More Information
When a block slides down an incline at constant velocity, the net work done on the system is zero. The work done by gravity is balanced by the work done against friction.
Tips
- Forgetting to include the direction of the frictional force when calculating work.
- Confusing normal force components on an incline, so it's important to consider only the perpendicular component.
