Problem 25 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A small block of mass \(m\) is given an initial speed \(v_{0}\) up a ramp inclined at angle \(\theta\) to the horizontal. It travels a distance \(d\) up the ramp and comes to rest. ( \(a\) ) Determine a formula for the coefficient of kinetic friction between block and ramp. (b) What can you say about the value of the coefficient of static friction?

Step-by-Step Solution

Answer

(a) \( \mu_k = \frac{\frac{1}{2} mv_0^2 - mgd \sin \theta}{mgd \cos \theta} \); (b) \( \mu_s > \mu_k \).

Step 1: Understand the forces involved

When the block moves up the ramp, there are several forces acting on it: gravitational force, normal force, and frictional force. The frictional force is what opposes the motion of the block as it goes up the ramp.

Step 2: Write the force equations

The block travels until it comes to rest, meaning the net work done on it is zero. The work done by the friction force and gravity is equal to the initial kinetic energy. The component of gravitational force along the plane is given by \(mg \sin \theta\) and the friction force is \( \mu_k N\), where \( \mu_k \) is the coefficient of kinetic friction.

Step 3: Express normal force

The normal force \(N\) is equal to the component of the gravitational force perpendicular to the plane, which is \(mg \cos \theta\).

Step 4: Apply energy conservation

The initial kinetic energy is given by \( \frac{1}{2} mv_0^2 \). As the block comes to rest, this is equal to the work done by both gravity and friction: \[ \frac{1}{2} mv_0^2 = mgd \sin \theta + \mu_k mgd \cos \theta \].

Step 5: Solve for the coefficient of kinetic friction

Rearrange the equation from the previous step to solve for \( \mu_k \): \[ \mu_k = \frac{\frac{1}{2} mv_0^2 - mgd \sin \theta}{mgd \cos \theta} \].

Step 6: Consider the static friction

The static friction coefficient \( \mu_s \) must be greater than the kinetic friction coefficient \( \mu_k \), because static friction always exceeds kinetic friction under static conditions, ensuring the block does not slide back down initially.

Key Concepts

Inclined Plane

An inclined plane is a flat surface tilted at an angle to the horizontal.
This simple concept is useful because it allows objects to travel vertically using less force than lifting straight up.
In the context of physics problems, an inclined plane might represent a ramp or a slope. When dealing with inclined planes, several important forces come into play:

• The gravitational force pulls the object downwards.
• The normal force is the perpendicular force exerted by the plane on the object.
• The frictional force resists the motion of the object along the plane.

Understanding these forces helps to solve many physics problems, such as finding how far an object might travel or how much it might slow down. On an incline, components of gravitational force need to be separated into parallel and perpendicular to the plane, which affects how we calculate these forces.

Normal Force

The normal force always acts perpendicular to the surface of contact.
On an inclined plane, it is crucial to remember that the normal force is not equal to the gravitational force.
Instead, it equals the component of the gravitational force that acts perpendicular to the plane.Given the angle of the incline, we can calculate the normal force using the formula:\[ N = mg \cos \theta \]Here,

• \( N \) is the normal force.
• The term \( mg \) is the weight of the object, where \( m \) is mass, and \( g \) is acceleration due to gravity.
• \( \cos \theta \) is the cosine of the angle of inclination.

Understanding the normal force is vital as it helps in computing friction. The normal force essentially acts as the counterbalance to the gravitational force's perpendicular component, keeping the object from passing through the surface.

Energy Conservation

Energy conservation is a fundamental principle in physics.
It states that energy cannot be created or destroyed, only transformed from one form to another.
When a block slides up an inclined plane, energy conservation helps determine how energy shifts between kinetic and potential forms.Initially, the block has kinetic energy given by:\[ \frac{1}{2} mv_0^2 \]As the block moves, this energy is converted into work done against friction and the gravitational component:\[ mgd \sin \theta + \mu_k mgd \cos \theta \]In this conservation of energy context:

• The initial kinetic energy is spent to overcome both the gravitational pull (up the slope) and the frictional force resisting motion.
• By solving the equation above, we find out how much energy is lost to these forces.
• Energy conservation ensures that the sum of input energies equals the sum of output energies in the isolated system.

Applying this principle allows us to calculate parameters like the coefficient of kinetic friction, shedding light on resistance forces active during the motion.