Problem 68 - University Physics with Modern Physics

Question

A 5.00 -kg package slides 1.50 \(\mathrm{m}\) down a long ramp that is inclined at \(24.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_{k}=0.310 .\) Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity; (c) the work done on the package by the normal force; (d) the total work done on the package. If the package has a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) at the top of the ramp, what is its speed after sliding 1.50 \(\mathrm{m}\) down the ramp?

Step-by-Step Solution

Answer

(a) -20.87 J, (b) 29.93 J, (c) 0 J, (d) 9.06 J; final speed is 2.91 m/s.

Step 1: Calculate the Force of Friction

The force of friction \( f_k \) is given by the formula \( f_k = \mu_k \cdot N \), where \( N \) is the normal force. The normal force \( N \) is given by \( N = mg \cos \theta \), where \( m = 5.00 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( \theta = 24.0^\circ \). Substitute the values to find \( N \):\[ N = 5.00 \times 9.81 \times \cos(24.0^\circ) = 44.88 \, \text{N} \]Then, calculate the force of friction:\[ f_k = 0.310 \times 44.88 = 13.91 \, \text{N} \]

Step 2: Work Done by Friction

The work done by friction \( W_f \) is given by \( W_f = -f_k \cdot d \), where \( d = 1.50 \, \text{m} \). The work is negative because friction opposes the motion:\[ W_f = -13.91 \times 1.50 = -20.87 \, \text{J} \]

Step 3: Work Done by Gravity

The work done by gravity \( W_g \) is calculated using \( W_g = mgd \sin \theta \):\[ W_g = 5.00 \times 9.81 \times 1.50 \times \sin(24.0^\circ) = 29.93 \, \text{J} \]

Step 4: Work Done by the Normal Force

The work done by the normal force \( W_n \) is zero because the force is perpendicular to the displacement:\[ W_n = 0 \, \text{J} \]

Step 5: Total Work Done on the Package

Calculate the total work done \( W_{total} \) by summing the work done by friction, gravity, and the normal force:\[ W_{total} = W_f + W_g + W_n = -20.87 + 29.93 + 0 = 9.06 \, \text{J} \]

Step 6: Final Speed of the Package

Use the work-energy principle to find the final speed \( v \):\[ \Delta K = W_{total} = \frac{1}{2}m(v^2 - u^2) \]where \( u = 2.20 \, \text{m/s} \) is the initial speed. Solve for \( v \):\[ 9.06 = \frac{1}{2} \times 5.00 \times (v^2 - (2.20)^2) \]\[ 9.06 = 2.5 \times (v^2 - 4.84) \]\[ v^2 - 4.84 = \frac{9.06}{2.5} \]\[ v^2 = 4.84 + 3.624 = 8.464 \]\[ v = \sqrt{8.464} = 2.91 \, \text{m/s} \]

Key Concepts

Kinetic Friction

Kinetic friction is a force that acts opposite to the direction of motion when two surfaces slide past each other. It arises due to the microscopic interactions between the surfaces in contact.
In our exercise, we have a package sliding down an inclined plane where kinetic friction comes into play.
Here, kinetic friction can be calculated using:

• A coefficient of kinetic friction \( \mu_{k} \), which is a constant value dependent on the materials in contact. For this problem, it is \( 0.310 \).
• The normal force \( N \), which is exerted perpendicular to the contact surface.

The force of friction \( f_k \) is obtained by multiplying the coefficient of kinetic friction \( \mu_{k} \) with the normal force \( N \) using the formula: \[ f_k = \mu_k \cdot N \] In this scenario, this value was found to be \( 13.91 \, \text{N} \). This frictional force acts backwards against the motion of the package, which is why the work done by kinetic friction is considered negative.

Inclined Plane

An inclined plane is a flat surface that is tilted at an angle to the horizontal. It allows for an object to be moved up or down the plane with less effort compared to lifting it directly upward.
In our exercise, the ramp forms an inclined plane that is set at an angle of \( 24.0^{\circ} \). This angle is crucial because it affects both the normal force and the gravitational force felt by the package.

• The inclined plane alters the direction of the gravitational force component acting parallel to the incline, facilitating the object's slide down the plane.
• It also directly influences the calculations for the normal and frictional forces as the forces need to be resolved into components parallel and perpendicular to the plane.

Understanding the inclined plane is essential because it changes how forces are applied to an object compared to forces felt on a horizontal plane.

Normal Force

The normal force is the support force exerted by a surface perpendicular to an object resting on it. It balances the component of the gravitational force acting perpendicular to the surface.
When an object, like our package, is on an inclined plane, the normal force \( N \) can be calculated using the formula: \[ N = mg \cos \theta \] Where \( m \) is the object's mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of inclination.
In this exercise, the normal force was calculated as \( 44.88 \, \text{N} \). Despite its magnitude, the normal force does no work on the package because it acts perpendicular to the direction of motion. This results in zero work contribution from the normal force.

• The primary role of the normal force in this context is to provide the necessary reaction for the frictional force to act.

Without it, kinetic friction could not occur.

Gravitational Force

Gravitational force is the attraction between two masses. For objects near the Earth's surface, this force pulls objects downwards. It is a crucial component in determining how objects move along inclined planes.
In this exercise, gravitational force can be decomposed into two components when dealing with inclined planes:

• One component acts parallel to the plane, causing the package to accelerate downward.
• The other component acts perpendicular to the plane, affecting the normal force.

To find the work done by gravity, the formula \( W_g = mgd \sin \theta \) is used, resulting in \( 29.93 \, \text{J} \) of work done in this scenario.
The gravitational force pulls the package down the ramp, promoting the movement and increasing the package's speed as it slides along the inclined surface.

• The gravitational force component parallel to the incline effectively aids in overcoming friction and contributing to the net work done on the package.

Understanding gravitational force in this context helps clarify how it governs the dynamics involved as objects interact with inclined surfaces.