Question

A boxed 10.0-kg computer monitor is dragged by friction 5.50 m upward along a conveyor belt inclined at an angle of 36.9${}^{\circ }$ above the horizontal. If the monitor's speed is a constant 2.10 cm/s, how much work is done on the monitor by (a) friction, (b) gravity, and (c) the normal force of the conveyor belt?

Short answer

Work by gravity: -3.3 kJ; by normal force: 0; friction typically adapts to zero net work.

Step-by-step solution

Identify the forces involved

The problem involves three main forces acting on the monitor: friction, gravity, and the normal force. We need to calculate the work done by each force as the monitor is dragged along the conveyor belt.

Calculate the work done by friction

The force due to friction ($F_{\text{friction}}$ ) can be calculated from the work-energy principle. Since the velocity is constant, net work is zero. The work done by friction, $W_{\text{friction}}$, is equal to the force of friction times the displacement $d$ times $\cos (\theta )$, where $\theta ={180}^{\circ }$ because friction opposes the motion: $$W_{\text{friction}}=F_{\text{friction}}\cdot d\cdot \cos ({180}^{\circ })=-F_{\text{friction}}\cdot 5.5.$$To solve for $F_{\text{friction}}$, we need additional information like the coefficient of friction; hence, typically it's derived from balance with other forces in practical scenarios.

Calculate the work done by gravity

The gravitational force acting on the monitor is $F_{\text{gravity}}=m\cdot g=10.0\cdot 9.8$. The work done by gravity is calculated by: $$W_{\text{gravity}}=F_{\text{gravity}}\cdot d\cdot \cos (\varphi ),$$where $\varphi ={36.9}^{\circ }+{90}^{\circ }={126.9}^{\circ }$. Thus, $$W_{\text{gravity}}=10\times 9.8\times 5.5\times \cos ({126.9}^{\circ }).$$Calculate to find the negative work done by gravity as it is acting downward.

Calculate the work done by the normal force

The normal force acts perpendicular to the direction of displacement, hence the angle between the direction of the normal force and the motion is ${90}^{\circ }$. Therefore, the work done by the normal force is:$$W_{\text{normal}}=F_{\text{normal}}\cdot d\cdot \cos ({90}^{\circ })=0.$$

Calculate numerical values

1. Compute the gravitational work: $$W_{\text{gravity}}=10\times 9.8\times 5.5\times \cos ({126.9}^{\circ })=-10.0\times 9.8\times 5.5\times (-0.6)=-3.3kJ.$$2. Work by normal force is already determined as 0.3. Work by friction is typically calculated to complement the zero net work in this situation.

Key concepts

Inclined Plane Problems

When dealing with inclined plane problems in physics, we often encounter scenarios where objects are moved along a slope or an incline. This exercise showcases such a problem, where we focus on a boxed monitor on a conveyor belt inclined at 36.9 degrees. The angle of the incline significantly impacts the forces acting on the object. Understanding how to break down these forces into components along the plane and perpendicular to it is key to solving these problems.

Key things to consider for inclined plane problems include:

• Identifying the incline angle: This affects the gravitational force component along the plane.
• Analyzing forces: Typically involve gravity, the normal force, and friction.
• Calculating displacement parallel to the incline, which helps determine work done by various forces.

For successful problem-solving, one should focus on decomposing forces into parallel and perpendicular components relative to the plane. This aids in calculating work done by each force more straightforwardly.

Frictional Force Calculations

Frictional force calculations are crucial in determining how much resistance an object encounters while moving along a surface. In this exercise, friction plays an important role as it opposes the motion of the monitor. The frictional force typically requires the coefficient of friction, which was not explicitly given, indicating the need for further context or information from experiments or balance equations.

Points to remember when calculating frictional forces include:

• Friction opposes motion, thus the work done is generally negative.
• Net work is zero when speed is constant, meaning frictional and other forces balance out.
• The work-energy principle can assist in identifying frictional forces by aligning with known energy changes.

Understanding these elements can help approach frictional calculations methodically, ensuring that forces such as friction are thoroughly accounted for in physics problems involving movement along inclined planes.

Gravitational Work

Gravitational work refers to the work done by gravitational force when an object moves along a particular path. In this scenario, the gravitational force is continuously acting downwards while the monitor is dragged upwards on the inclined conveyor belt.

Consider the following when calculating gravitational work:

• The gravitational force ($F_{\text{gravity}}=m\cdot g$) depends on the mass and gravitational acceleration.
• The angle affects the cosine component in the work equation, where$\cos ({126.9}^{\circ })$ indicates the direction opposing movement.
• Negative work illustrates gravity acting against the upward motion.

Using the correct angle references and understanding the impact of gravity in different motion scenarios, you can effectively compute the gravitational work in similar inclined plane problems. The correct interpretation of direction and angles is vital in minimizing errors in these calculations.