Question

A 1.00 -kg block is pushed up and down a rough plank of length \(L=2.00 \mathrm{~m},\) inclined at \(30.0^{\circ}\) above the horizontal. From the bottom, it is pushed a distance \(L / 2\) up the plank, then pushed back down a distance \(L / 4\), and finally pushed back up the plank until it reaches the top end. If the coefficient of kinetic friction between the block and plank is 0.300 , determine the work done by the block against friction.

Short answer

Answer: The total work done against friction in this scenario is 7.66 J.

Step-by-step solution

Calculate the force of friction

First, we need to find the force of friction acting on the block. The frictional force can be calculated using the formula: $$F_{friction} = \mu F_N$$ where \(\mu\) is the coefficient of kinetic friction and \(F_N\) is the normal force acting on the block. As the inclined plane is at an angle of \(30.0^\circ\), the normal force can be calculated as: $$F_N = mg\cos(\theta)$$ Substituting the given values, we have $$F_N = (1.00\,\text{kg})(9.81\,\text{m/s}^2)\cos(30.0^\circ)$$ After calculating this, we get the normal force \(F_N = 8.50\,\text{N}\) Now, we can calculate the force of friction using the coeffcient of kinetic friction: $$F_{friction} = 0.300 \times 8.50\,\text{N}$$ This results in a force of friction \(F_{friction} = 2.55\,\text{N}\)

Calculate the work done against friction

The work done against friction can be calculated by multiplying the force of friction with the distance covered, as follows: $$W = F_{friction} d$$ To find the total work done, we need to calculate the work done when the block is pushed up the inclined plane, pushed down the inclined plane and pushed back up the inclined plane to the top end. 1. Pushed a distance \(L/2\) up the inclined plane In this case, the distance covered is \(d_1 = (2.00\,\text{m})/2 = 1.00\,\text{m}\) The work done against friction is: $$W_1 = (2.55\,\text{N})(1.00\,\text{m}) = 2.55\,\text{J}$$ 2. Pushed a distance \(L/4\) down the inclined plane In this case, the distance covered is \(d_2 = (2.00\,\text{m})/4 = 0.50\,\text{m}\) The work done against friction is: $$W_2 = (2.55\,\text{N})(0.50\,\text{m}) = 1.28\,\text{J}$$ 3. Pushed back up the inclined plane until it reaches the top end In this case, the distance covered is \(d_3 = (2.00\,\text{m})/2 + (2.00\,\text{m})/4 = 1.50\,\text{m}\) The work done against friction is: $$W_3 = (2.55\,\text{N})(1.50\,\text{m}) = 3.83\,\text{J}$$

Sum the total work done against friction

Finally, we can calculate the total work done against friction by summing the work done in all three cases: Total work done = \(W_1 + W_2 + W_3 = 2.55\,\text{J} + 1.28\,\text{J} + 3.83\,\text{J} = 7.66\,\text{J}\)

Key concepts

Kinetic Friction

Kinetic friction is the force that opposes the relative motion between two surfaces in contact when they are moving past each other. This force is always in the direction opposite to the direction of motion and is a crucial factor when calculating the work done against friction.

In the given exercise, the kinetic friction is the force the block must overcome while being pushed up and down the inclined plane. The coefficient of kinetic friction, denoted by \(\mu\), is a measure of how much frictional force is produced for a given normal force and does not depend on the area of contact or the speed of the sliding object. Here, \(\mu = 0.300\) indicates a medium amount of friction between the block and the plank.

Inclined Plane Physics

An inclined plane is a flat surface tilted at an angle to the horizontal. Physics involving inclined planes requires consideration of gravity, normal force, and friction acting at different angles compared to flat surfaces.

In our exercise, the 30-degree inclination changes how the force of gravity is distributed: into components parallel and perpendicular to the plane's surface. The component perpendicular to the inclined plane determines the normal force, which is vital for calculating the frictional force. This particular setup demands a clear understanding of vector resolution to solve problems effectively.

Normal Force Calculation

The normal force is the force exerted by a surface that supports the weight of an object resting on it. On an inclined plane, the normal force is less than the object's weight and can be computed as \(mg\cos(\theta)\), where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(\theta\) is the incline angle.

For the exercise, the normal force calculation is crucial as it directly impacts the amount of friction the block encounters. With the mass of the block and the incline angle known, the normal force was determined to be \(8.50\,\text{N}\), which consequently affected the kinetic friction force and the overall work done against it during the motion of the block.

Work-Energy Principle

The work-energy principle states that work done on an object is equal to the change in its kinetic energy. Calculating work done against friction is an application of this principle, as friction converts kinetic energy into thermal energy, effectively reducing the block's kinetic energy.

In the exercise, to determine the total work done against friction, we calculate the work for each segment of the block's path separately: up, down, and back up the plane. This is then summed to get the total work done. The work done against friction on the inclined plane not only helps understand the energy needed to move an object but also has practical implications in fields like engineering and transportation.