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A \(46-\mathrm{kg}\) skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of \(-1.2 \mathrm{m} / \mathrm{s}\). Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude as, but opposite direction to, the force that the wall applies to her).

Short Answer

Expert verified
The skater exerts a force of 69 N on the wall in the direction away from it.

Step by step solution

01

Understand the Problem

The problem requires us to find the magnitude and direction of the average force that the skater exerts on the wall. This force is the same magnitude but in opposite direction to the force exerted by the wall on the skater.
02

Identify the Known Values

The mass of the skater, \(m = 46 \, \text{kg}\), velocity \(v = -1.2 \, \text{m/s}\), and the time of contact, \(t = 0.80 \, \text{s}\) are given.
03

Use the Impulse-Momentum Theorem

The impulse experienced by an object is equal to the change in its momentum, which can be expressed as \(F_{\text{avg}} \cdot \Delta t = m \cdot \Delta v\). Here, \(\Delta v = -1.2 \, \text{m/s} - 0 \, \text{m/s} = -1.2 \, \text{m/s}\).
04

Rearrange the Formula to Solve for the Average Force

Rearrange the impulse-momentum theorem to solve for the force: \( F_{\text{avg}} = \frac{m \cdot \Delta v}{\Delta t} \).
05

Substitute the Values into the Equation

Substitute the known values into the equation: \( F_{\text{avg}} = \frac{46 \, \text{kg} \cdot (-1.2 \, \text{m/s})}{0.80 \, \text{s}} = -69 \, \text{N}\).
06

Determine the Direction of the Force

The negative sign indicates that the direction of the average force exerted by the skater on the wall is opposite to her velocity. Hence, she pushes in the direction away from the wall.

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Key Concepts

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

Average Force
When discussing the average force, think of it as the constant force that, if applied over a certain time period, equals the force calculated during that time. It helps us understand how an impact spreads its effects over time.
In the context of the skater pushing against the wall, the average force (\( F_{\text{avg}} \)) is what she applies to the wall during her push. It combines her mass and velocity change with the time she maintains contact with the wall.
To find the average force, we use the formula:
\[F_{\text{avg}} = \frac{m \cdot \Delta v}{\Delta t}\]
where:
  • \( m \) is the mass of the skater,
  • \( \Delta v \) is the change in velocity, and
  • \( \Delta t \) is the time of contact.
Using this formula, the average force gives insights into how her mass and change in velocity affect the force exerted over the duration of her push.
Momentum
Momentum is a measure of motion that combines both an object's mass and velocity. It gives us an idea of how hard it is to stop a moving object.
The formula for momentum (\( p \)) is expressed as:\[p = m \cdot v\]In our example, the skater initially has zero momentum because she starts standing still. However, when she pushes against the wall, her velocity changes, which in turn changes her momentum.
The impulse-momentum theorem states that the change in momentum is equal to the impulse applied to the object. Impulse, like force applied over time, alters an object’s momentum. With the skater’s velocity changing from 0 to \(-1.2 \, \text{m/s}\), you can see the change in momentum as:\[\Delta p = m \cdot \Delta v = 46 \, \text{kg} \cdot (-1.2 \, \text{m/s})\]This numerical change in momentum is useful for understanding how pushing against a wall, or any force over time, changes an object’s motion.
Velocity Change
Understanding velocity change is crucial in scenarios involving motion, especially when an object starts from rest and gains speed in another direction.
In this exercise, the skater goes from being still to moving backward at \(-1.2 \, \text{m/s}\). This change in velocity (\( \Delta v \)) is vital because it reflects how the skater's push against the wall altered her state of motion.
To calculate the velocity change, we use:\[\Delta v = v_f - v_i\]where:
  • \( v_f \) is the final velocity, and
  • \( v_i \) is the initial velocity.
For the skater, we go from \( 0 \, \text{m/s}\) to \(-1.2 \, \text{m/s}\), yielding a \( \Delta v \) of \(-1.2 \, \text{m/s}\). This negative sign denotes a backward motion, opposite to any initial forward direction. Understanding these changes helps tie together concepts of force, impulse, and the resulting movement.

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Most popular questions from this chapter

Starting with an initial speed of \(5.00 \mathrm{m} / \mathrm{s}\) at a height of \(0.300 \mathrm{m}\), a 1.50-kg ball swings downward and strikes a 4.60-kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

A student \((m=63 \mathrm{kg})\) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of \(0.040 \mathrm{s} .\) The average force exerted on him by the ground is \(+18000 \mathrm{N},\) where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

A basketball \((m=0.60 \mathrm{kg})\) is dropped from rest. Just before striking the floor, the ball has a momentum whose magnitude is \(3.1 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) At what height was the basketball dropped?

One object is at rest, and another is moving. The two collide in a one- dimensional, completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. The speed of the object that is moving initially is \(25 \mathrm{m} / \mathrm{s} .\) The masses of the two objects are 3.0 and \(8.0 \mathrm{kg} .\) Determine the final speed of the two-object system after the collision for the case when the large-mass object is the one moving initially and the case when the small-mass object is the one moving initially.

An astronaut in his space suit and with a propulsion unit (empty of its gas propellant) strapped to his back has a mass of 146 kg. The astronaut begins a space walk at rest, with a completely filled propulsion unit. During the space walk, the unit ejects some gas with a velocity of \(+32 \mathrm{m} / \mathrm{s}\). As a result, the astronaut recoils with a velocity of \(-0.39 \mathrm{m} / \mathrm{s}\). After the gas is ejected, the mass of the astronaut (now wearing a partially empty propulsion unit) is 165 kg. What percentage of the gas was ejected from the completely filled propulsion unit?

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