Question

A block of mass \(10 \mathrm{~kg}\) moves along a surface inclined \(30^{\circ}\) relative to the horizontal. The center of gravity of the block is elevated by \(3.0 \mathrm{~m}\) and the kinetic energy of the block decreases by \(50 \mathrm{~J}\). The block is acted upon by a constant force \(\mathbf{R}\) parallel to the incline and by the force of gravity. Assume frictionless surfaces and let \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\). Determine the magnitude and direction of the constant force \(\mathbf{R}\), in \(\mathrm{N}\).

Short answer

The magnitude of the force \(\mathbf{R}\) is 40.72 N, and it acts opposite to the motion of the block.

Step-by-step solution

Understand the problem

Identify that the block is moving on a frictionless inclined plane with a decrease in kinetic energy and elevation in its center of gravity. The goal is to find the constant force \(\mathbf{R}\) acting parallel to the incline.

Calculate the change in potential energy

The change in potential energy is given by \(mgh\), where \(m = 10 \,kg\), \(g = 9.81 \, m/s^2\), and \(h = 3.0 \, m\). \ \[ \Delta U = mgh = 10 \, kg \cdot 9.81 \, m/s^2 \cdot 3.0 \, m = 294.3 \, J \]

Relate kinetic and potential energy changes

The work done by the force \(\mathbf{R}\) on the block changes its kinetic energy and potential energy. Therefore, \ \[W = \Delta KE + \Delta U\] where \(\Delta KE = -50 \, J\) and \(\Delta U = 294.3 \, J\). So, \ \[W = -50 \, J + 294.3 \, J = 244.3 \, J\]

Find the displacement along the incline

With the height of 3.0 m and the incline angle \(30^{\circ}\), use trigonometry: \ \[ h = L \sin 30^{\circ} \Rightarrow L = \frac{h}{\sin 30^{\circ}} = \frac{3.0 \, m}{0.5} = 6.0 \, m \]

Calculate the force \(\mathbf{R}\)

Use the work-energy principle, \ \[W = R \cdot L \Rightarrow R = \frac{W}{L} = \frac{244.3 \, J}{6.0 \, m} = 40.72 \, N \]

Determine the direction of the force

Since the kinetic energy decreases, the force \(\mathbf{R}\) must be acting opposite to the motion of the block to cause a decrease in speed.

Key concepts

Potential Energy

Potential energy is the energy that is stored in an object due to its position relative to some zero position. In this exercise, the block of mass is elevated by 3.0 meters. The elevation results in a change in potential energy, which can be calculated using the formula \(\text{Potential Energy} = mgh\). Here, \(m\) is the mass of the block (10 kg), \(g\) is the acceleration due to gravity (9.81 m/s²), and \(h\) is the height (3.0 m). This relationship tells us how the mass and height affect the potential energy. For our block, the potential energy change is calculated as: \[ \Delta U = mgh = 10 \,kg \cdot 9.81 \, m/s^2 \cdot 3.0 \,m = 294.3 \,J \]. This is the energy the block gains by being elevated. The concept of potential energy is crucial since it shows how the block's position affects its energy state.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. The formula to calculate kinetic energy is \(\text{Kinetic Energy} = \frac{1}{2} mv^2\), where \(m\) is the mass of the object and \(v\) is its velocity. In our inclined plane problem, the kinetic energy of the block decreases by 50 J. This means the block is losing energy because it is slowing down. We represent this change as \[ \Delta KE = -50 \, J \]. Understanding how kinetic energy changes are essential in energy conservation problems. It helps us see the energy transformation as the block moves up the incline and how external forces impact its speed.

Work-Energy Principle

The work-energy principle ties the concepts of work and energy together. It states that the work done by all forces acting on an object equals the change in its kinetic energy. In equation form, \[ W = \Delta KE \]. In our inclined plane problem, this principle is extended to include the change in potential energy as well. The total work done by the force \( \mathbf{R} \) is given by the sum of the change in kinetic and potential energies: \[ W = \Delta KE + \Delta U \]. Substituting the given values we get \[ W = -50 \, J + 294.3 \, J = 244.3 \, J \]. This tells us that the work done by the force is equal to 244.3 J. The work-energy principle is a powerful tool in physics because it simplifies the analysis of complex systems by focusing on energy changes.

Trigonometry in Physics

Trigonometry plays a significant role in solving physics problems involving inclined planes. We use trigonometry to relate the geometric properties of the incline to the physical quantities. In this problem, we need to find the displacement along the incline. Given the height of 3.0 meters and an angle of 30°, we use the sine function: \[ h = L \sin 30^{\circ} \rightarrow L = \frac{h}{\sin 30^{\circ}} = \frac{3.0 \, m}{0.5} = 6.0 \, m \]. Here, \( L \) is the length of the incline. Once we have \( L \), we can then use it to find the force \( \mathbf{R} \) using the work-energy principle. This step is crucial as it connects the height of the block's elevation to the total distance it travels along the incline. Understanding trigonometry allows us to break down forces and movements into manageable components.