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A young ice skater with mass \(40.0 \mathrm{~kg}\) has fallen and is sliding on the frictionless ice of a skating rink with a spced of \(20.0 \mathrm{~m} / \mathrm{s}\). (a) What is the magnitude of her linear momentum when she has this speed? (b) What is her kinetic energy? (c) What constant net horizontal force must be applied to the skater to bring her to rest in \(5.00 \mathrm{~s}\) ?

Short Answer

Expert verified
The linear momentum is \(800.0 \mathrm{~kg~m/s}\), the kinetic energy is \(8000.0 \mathrm{~J}\), and the constant net force needed to bring the ice skater to rest is \(160.0 \mathrm{~N}\)

Step by step solution

01

Calculate the Linear Momentum

Linear momentum (p) is calculated by multiplying the mass (m) of the object by its velocity (v). Using the equation p=mv, substitute the given values into the equation to find the solution: p = 40.0 kg * 20.0 m/s.
02

Calculate the Kinetic Energy

Kinetic energy (K.E) is given by the equation K.E = 0.5*m*v^2. Plug in the given values into the equation: K.E = 0.5 * 40.0 kg * (20.0 m/s)^2.
03

Find the constant net horizontal force

The net force (f) required to stop the skater is given by f=ma. The acceleration (a) to used here is the change in velocity over the time taken to stop. Given that the ice skater comes to a halt, the final velocity is zero. The initial velocity is the speed of the skater and the time taken to stop is given. Substitute these values to get the acceleration: a = (final velocity - initial velocity) / time = (0 - 20.0 m/s) / 5.0 s. Then plug in the mass and acceleration into the force equation to get the solution: f= 40.0 kg * a.

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Key Concepts

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

Kinetic Energy Calculation
Kinetic energy is a measure of the energy an object possesses due to its motion. It’s a fundamental concept in physics, essentially quantifying how much work an object can perform due to its velocity. The formula for calculating kinetic energy (K.E) is \(K.E = \frac{1}{2}mv^2\), where \(m\) is the mass of the object and \(v\) is its velocity.

For our young ice skater with a mass of \(40.0\,\text{kg}\) gliding across the ice at a speed of \(20.0\,\text{m/s}\), we see kinetic energy in action. By plugging these values into the kinetic energy equation, we obtain \(K.E = \frac{1}{2} \times 40.0\,\text{kg} \times (20.0\,\text{m/s})^2\), which gives us the total kinetic energy that the skater has while moving. Understanding kinetic energy is essential not just for physics problems like these but also in everyday scenarios, like assessing the impact force of a moving vehicle.
Net Force Calculation
Net force is instrumental in understanding how an object's motion changes over time. Newton’s second law of motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (\(F = ma\)). This principle forms the basis for our calculation of the net force needed to stop the ice skater.

For the skater scenario, we need to compute the force required to bring her to rest from her current velocity over a specific period. With an initial velocity of \(20.0\,\text{m/s}\) and a stopping time of \(5.00\,\text{s}\), we calculate the deceleration required. We then use the mass of the skater, alongside this deceleration, to find the net force. In our case, the negative acceleration indicates a force applied in the direction opposite to the skater's motion, consistent with what is needed to halt her movement.
Physics Problem Solving
Solving physics problems effectively involves understanding the concepts at play, identifying the known variables, and then methodically applying the appropriate physics equations. A step-by-step approach helps to dissect the problem and manage it in smaller, more manageable parts.

For instance, with the ice skater’s problem, it’s crucial to recognize the relevance of linear momentum, kinetic energy, and net force. After writing down what’s given, such as mass and velocity, and what needs to be found like linear momentum, kinetic energy, and the force required to stop the skater, we can tackle each part sequentially using formulas. It can be helpful to sketch the scenario, which may provide visual clues and aid in understanding the physics concepts related to the problem. Furthermore, always double-check calculations for accuracy and ensure that the units are consistent throughout the problem-solving process.

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Most popular questions from this chapter

A 5.00gbullet is fired horizontally into a \(1.20 \mathrm{~kg}\) wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20 . The bullet remains embedded in the block, which is observed to slide \(0.310 \mathrm{~m}\) along the surface before stopping. What was the initial speed of the bullet?

In a game of phy sics and skill, a rigid block (A) with mass \(m\) sits at rest at the edge of a frictionless air table, \(1.20 \mathrm{~m}\) above the floor. You slide an identical block \((B)\) with initial speed \(v_{B}\) toward \(A\). The blocks have a head-on elastic collision, and block \(A\) leaves the table with a horizontal velocity. The goal of the game is to have block \(A\) land on a target on the floor. The targct is a horizontal distance of \(2.00 \mathrm{~m}\) from the edge of the table. What is the initial speed \(v_{n}\) that accomplishes this? Neglect air resistance.

An apple with mass \(M\) is hanging at rest from the lower end of a light vertical rope. A dart of mass \(M / 4\) is shot vertically upward, strikes the bottom of the apple, and remains cmbedded in it. If the specd of the dart is \(\mathrm{c}_{0}\) just before it strikes the apple, how high docs the apple move upward because of its collision with the dart?

A Ricocheting Bullet. A \(0.100 \mathrm{~kg}\) stone rests on a frictionless, horizontal surface. A bullet of mass \(6.00 \mathrm{~g}\). traveling horizontally at \(350 \mathrm{~m} / \mathrm{s}\), strikes the stone and rebounds horizontally at right angles to its original direction with a speed of \(250 \mathrm{~m} / \mathrm{s}\). (a) Compute the magnitude and direction of the velocity of the stonc after it is struck. (b) Is the collision perfectly clastic?

A10.0g marble slides to the left at a speed of \(0.400 \mathrm{~m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, clastic collision with a larger \(30.0 \mathrm{~g}\) marble sliding to the right at speed of \(0.200 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{E} 8.48\) ). (a) Find the velocity of each marble (magnitude and dircction) after the collision. (since the collision is head-on, all motion is along a line. (b) Calculate the change in momentum (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the change in kinetic energy (the kinctic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

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