Problem 8 - College Physics

Question

A \(\mathrm{A} 68.5 \mathrm{kg}\) skater moving initially at 2.40 \(\mathrm{m} / \mathrm{s}\) on rough horizontal ice comes to rest uniformly in 3.52 s due to friction from the ice. What force does friction exert on the skater?

Step-by-Step Solution

Answer

The force exerted by friction is \(-46.7 \, \text{N}\).

Step 1: Understanding the Problem

We need to calculate the force exerted by friction on the skater. We know the skater's initial velocity, final velocity, skater's mass, and the time taken for the skater to come to a stop.

Step 2: Identify Given Values

From the problem, we have the skater's initial velocity \(v_i = 2.40 \, \text{m/s}\), final velocity \(v_f = 0 \, \text{m/s}\), mass \(m = 68.5 \, \text{kg}\), and time \(t = 3.52 \, \text{s}\).

Step 3: Determine Acceleration

Use the formula for acceleration \(a = \frac{v_f - v_i}{t}\). Substitute the given values: \(a = \frac{0 - 2.40}{3.52}\). This results in \(a = -0.682 \, \text{m/s}^2\).

Step 4: Calculate Force

Use Newton's second law, \(F = m \cdot a\). Substitute the known values: \(F = 68.5 \cdot (-0.682)\). This gives the force \(F = -46.7 \, \text{N}\).

Step 5: Interpret the Result

The negative sign of the force indicates that it opposes the motion of the skater, meaning friction acts opposite to the direction of movement.

Key Concepts

Friction

Friction is a force that opposes the relative motion between two surfaces in contact. In this problem, the skater is moving on rough ice, which means there is significant friction at play. Friction acts to slow down objects sliding over each other, and it always acts in the opposite direction to the motion.

There are two main types of friction: static and kinetic. Static friction keeps an object at rest, while kinetic friction acts on moving objects. In this case, once the skater starts moving and continues to slide to a stop, kinetic friction is the key factor. It is responsible for bringing the skater to a halt.

• Friction depends on the nature of the surfaces and the force pressing them together.
• In the skater's case, the rough ice creates more friction than smooth ice would.

Newton's Second Law

Newton's second law of motion provides a way to understand how forces influence motion. It states that the force applied to an object equals the mass of the object multiplied by its acceleration: \( F = m \cdot a \). This formula forms the backbone of many calculations in physics.

In the scenario with the skater, Newton's second law helps us calculate the friction force. By understanding how the skater's velocity changes due to friction, we can determine the skater's acceleration. Once we have the acceleration, we multiply it by the skater's mass to find the force exerted by friction.

• This law shows the direct relationship between force, mass, and acceleration.
• It's crucial for predicting how an object will move when acted upon by various forces.

Acceleration

Acceleration is the rate of change of velocity of an object. When an object speeds up or slows down, it is accelerating. In this problem, the skater initially moves at 2.40 m/s and comes to a stop, meaning there's a change in velocity.

The skater's acceleration can be calculated using the formula: \[ a = \frac{v_f - v_i}{t} \]

• Inserting the skater's initial (\( v_i = 2.40 \, \text{m/s} \)) and final velocities (\( v_f = 0 \, \text{m/s} \)), along with time (\( t = 3.52 \, \text{s} \)), gives us the value of acceleration as \( -0.682 \, \text{m/s}^2 \).
• The negative sign signifies a decrease in speed, which is expected since the skater is slowing down.

Force Calculation

Force calculation is at the heart of solving problems involving motion and is often carried out using Newton's laws. Here, using Newton's second law, we calculate the force exerted by friction on the skater:

\[ F = m \cdot a \]
Substituting the given values (\( m = 68.5 \, \text{kg} \), \( a = -0.682 \, \text{m/s}^2 \)), we get the force as:\[ F = 68.5 \cdot (-0.682) = -46.7 \, \text{N} \]

• The negative sign of the force indicates that it acts in the opposite direction to the skater's motion.
• This friction force is what ultimately brings the skater to rest.