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

Short Answer

Expert verified
The friction force exerted on the skater is $-46.58N$. The negative sign indicates that the force is exerted in the direction opposite to the skater's initial motion.

Step by step solution

01

Calculate the Acceleration

We know that initial velocity \(u = 2.40 \mathrm{~m/s}\), final velocity \(v = 0 \mathrm{~m/s}\) (as skater comes to rest) and time \(t = 3.52 \mathrm{~s}\). We can substitute these values in the formula for acceleration, which is \(a = \frac{\Delta v}{\Delta t}\). Therefore, acceleration \(a = \frac{v - u}{t} = \frac{0 - 2.40}{3.52} = -0.68 \mathrm{~m/s^2}\). The negative sign indicates that the acceleration is in the opposite direction to the initial velocity.
02

Find the Friction Force

We have mass \(m = 68.5 \mathrm{~kg}\) and acceleration \(a = -0.68 \mathrm{~m/s^2}\). We can use these in the formula \(F = ma\) to find the force. This gives us \(F = 68.5 \times -0.68 = -46.58 N\). The negative sign again indicates the backward direction, which means the force exerted by friction is opposite to the skater's initial motion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Deceleration
When an object slows down at a steady rate, we say it is experiencing uniform deceleration. In the context of our skater, uniform deceleration means that the skater's decrease in speed over time is constant. This constant rate of slowing down can be calculated by taking the change in velocity over the time period in which the deceleration occurs.

To visualize this concept, think of the skater moving on ice: initially, they glide smoothly, then friction starts to act against the movement and bit by bit, the skater's speed reduces until they come to a complete stop. The ease with which we can calculate the level of deceleration is due to its uniform nature. If the rate of deceleration varied over time, the calculations would become more complex.
Acceleration Formula
The acceleration formula is key to understanding how objects speed up or slow down. It is defined as the rate of change of velocity per unit time. Mathematically, this is expressed as: \[ a = \frac{\Delta v}{\Delta t} \]
where \( a \) is the acceleration, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the change in time. For the skater, we see this in action as we compute the deceleration, which is a form of acceleration in a direction opposite to the initial movement.

It’s crucial to understand that acceleration isn't just about speeding up; it also encompasses slowing down, which is often referred to as 'negative acceleration' or 'deceleration'. In our skater's case, the negative acceleration indicates that the skater is slowing down rather than speeding up.
Newton's Second Law
At the heart of motion and forces is Newton's second law of motion, often expressed as the simple formula: \[ F = ma \], where \( F \) represents force, \( m \) is mass, and \( a \) is acceleration. This law tells us that the force acting on an object is equal to the mass of the object multiplied by its acceleration.

In practical terms for our scenario, the skater's mass doesn't change, but the friction force imposes an acceleration (in this case, deceleration) on the skater. Newton's second law allows us to calculate the exact force of friction by multiplying the mass of the skater by their rate of deceleration. This is how we determined the friction force acting against the skater on the ice. It is quintessential to recognize that force and acceleration are vector quantities, which means they have both magnitudes and direction, as highlighted by the negative signs in our calculations.

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Most popular questions from this chapter

The position of a training helicopter (weight \(2.75 \times 10^{5} \mathrm{~N}\) ) in a test is given by \(\hat{r}=\left(0.020 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} \hat{\imath}+\) \((2.2 \mathrm{~m} / \mathrm{s}) t \hat{\jmath}-\left(0.060 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} \hat{k} .\) Find the net force on the helicopter at \(t=5.0 \mathrm{~s}\)

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