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A 68.5-kg skater moving initially at 2.40 m/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?

Short Answer

Expert verified
The friction force is approximately -46.7 N, acting opposite to the motion.

Step by step solution

01

Understand the Problem

We are given the mass of a skater, the initial velocity, the final velocity, and the time taken to come to rest. The objective is to find the force of friction exerted on the skater, which will require using principles of mechanics like velocity, acceleration, and force.
02

Identify Known Values

The known values are the initial velocity of the skater \(v_i = 2.40 \text{ m/s}\), the final velocity \(v_f = 0 \text{ m/s}\), the time taken \(t = 3.52 \text{ s}\), and the mass of the skater \(m = 68.5 \text{ kg}\).
03

Find Deceleration

Since the skater comes to a rest uniformly, we use the equation of motion: \(v_f = v_i + at\). Substitute \(v_f = 0\), \(v_i = 2.40 \text{ m/s} \), and \(t = 3.52 \text{ s} \) to solve for acceleration \(a\). \(0 = 2.40 + a \times 3.52\), thus \(a = -\frac{2.40}{3.52}\approx -0.682\) m/s².
04

Apply Newton's Second Law

Using Newton's Second Law \( F = ma \), substitute \(m = 68.5 \text{ kg} \) and \(a = -0.682 \text{ m/s}^2\). Therefore, \( F = 68.5 \times (-0.682) \approx -46.7 \text{ N} \). The negative sign indicates that the force is acting opposite to the motion.

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Key Concepts

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

Understanding Newton's Second Law
Newton's Second Law of Motion is a fundamental principle in physics that relates the force applied to an object to its mass and the acceleration it experiences. The law is often expressed with the famous formula:
  • \( F = ma \)
where:
  • \( F \) is the net force applied to the object, measured in Newtons (N)
  • \( m \) is the mass of the object in kilograms (kg)
  • \( a \) is the acceleration in meters per second squared (\( \text{m/s}^2 \))
In the context of the exercise, the force exerted by friction acts against the skater's motion, causing them to slow down and eventually stop. This deceleration is due to the negative acceleration resulting from the force of friction. By measuring the skater’s mass and the acceleration, we can determine the amount of force required to bring the skater to a stop over a given period.
Exploring Deceleration
Deceleration is simply negative acceleration, meaning it is the process of slowing down. It's quantified with the same units as acceleration, \( \text{m/s}^2 \), but it acts in the direction opposite to the motion.When calculating deceleration, we use the formula:
  • \( a = \frac{v_f - v_i}{t} \)
where:
  • \( v_f \) is the final velocity
  • \( v_i \) is the initial velocity
  • \( t \) is the time interval over which the change occurs
In the exercise example, the skater is initially moving at 2.40 m/s and comes to rest, so the final velocity is 0. By inserting these values into the deceleration formula, we understand how quickly they slow down. The deceleration value is crucial for determining how strong the frictional force needs to be to stop the skater within the specified time.
Understanding Uniform Motion
Uniform motion refers to motion at a constant speed in a straight line. It's characterized by having both a constant velocity and zero acceleration. In our case, the skater starts in uniform motion, gliding smoothly over the ice until the frictional force comes into play, causing deceleration. If there were no friction, and thus no external forces acting on the skater, they would continue in uniform motion indefinitely according to Newton's First Law of Motion (law of inertia). However, because friction acts in the opposite direction, it breaks this uniform motion. It causes the skater to experience a uniform deceleration, bringing them to a stop. This transition from uniform motion to resting illustrates how external forces and uniform motion concepts work together in practical situations.

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Most popular questions from this chapter

A loaded elevator with very worn cables has a total mass of 2200 kg, and the cables can withstand a maximum tension of 28,000 N. (a) Draw the free-body force diagram for the elevator. In terms of the forces on your diagram, what is the net force on the elevator? Apply Newton's second law to the elevator and find the maximum upward acceleration for the elevator if the cables are not to break. (b) What would be the answer to part (a) if the elevator were on the moon, where \(g =\) 1.62 m/s\(^2\)?

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

A small 8.00-kg rocket burns fuel that exerts a timevarying upward force on the rocket (assume constant mass) as the rocket moves upward from the launch pad. This force obeys the equation \(F = A + Bt^2\). Measurements show that at \(t\) = 0, the force is 100.0 N, and at the end of the first 2.00 s, it is 150.0 N. (a) Find the constants \(A\) and \(B\), including their SI units. (b) Find the \(net\) force on this rocket and its acceleration (i) the instant after the fuel ignites and (ii) 3.00 s after the fuel ignites. (c) Suppose that you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.00 s after fuel ignition?

Two dogs pull horizontally on ropes attached to a post; the angle between the ropes is 60.0\(^\circ\). If Rover exerts a force of 270 N and Fido exerts a force of 300 N, find the magnitude of the resultant force and the angle it makes with Rover's rope.

Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (\(\textbf{Fig. P4.39}\)). A woman wearing golf shoes (for traction) pulls horizontally on the 6.00-kg crate with a force \(F\) that gives the crate an acceleration of 2.50 m/s\(^2\). (a) What is the acceleration of the 4.00-kg crate? (b) Draw a free-body diagram for the 4.00-kg crate. Use that diagram and Newton's second law to find the tension \(T\) in the rope that connects the two crates. (c) Draw a free-body diagram for the 6.00-kg crate. What is the direction of the net force on the 6.00-kg crate? Which is larger in magnitude, \(T\) or \(F\)? (d) Use part (c) and Newton's second law to calculate the magnitude of \(F\).

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