/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 A 63 -kg skier coasts up a snow-... [FREE SOLUTION] | 91Ó°ÊÓ

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A 63 -kg skier coasts up a snow-covered hill that makes an angle of \(25^{\circ}\) with the horizontal. The initial speed of the skier is \(6.6 \mathrm{~m} / \mathrm{s}\). After coasting a distance of \(1.9 \mathrm{~m}\) up the slope, the speed of the skier is \(4.4 \mathrm{~m} / \mathrm{s}\). (a) Find the work done by the kinetic frictional force that acts on the skis. (b) What is the magnitude of the kinetic frictional force?

Short Answer

Expert verified
(a) The work done by friction is approximately -1284.1 J. (b) The magnitude of the kinetic frictional force is 676.9 N.

Step by step solution

01

Identify Given Data

We have a skier starting with an initial speed \( v_i = 6.6 \, \text{m/s} \) and final speed \( v_f = 4.4 \, \text{m/s} \). The skier moves up a slope 1.9 meters long, inclined at an angle \( \theta = 25^{\circ} \). The mass of the skier is \( m = 63 \, \text{kg} \). We need to find the work done by kinetic friction \( W_\text{friction} \) and its magnitude.
02

Calculate Change in Gravitational Potential Energy

The change in gravitational potential energy \( \Delta U \) can be calculated using \( \Delta U = mgh \), where \( h \) is the vertical height: \( h = 1.9 \times \sin(25^{\circ}) \). First, calculate \( h \), then \( \Delta U = (63)(9.8)(1.9 \times \sin(25^{\circ})) \approx 521.8 \, \text{J} \).
03

Calculate Change in Kinetic Energy

The change in kinetic energy \( \Delta KE \) is given by \( \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) \). Substitute the values: \( \Delta KE = \frac{1}{2}(63)((4.4)^2 - (6.6)^2) = \frac{1}{2}(63)(19.36 - 43.56) = \frac{1}{2}(63)(-24.2) \approx -762.3 \, \text{J} \).
04

Use Work-Energy Principle

According to the work-energy principle, the total work done \( W_\text{total} \) is equal to the change in kinetic energy \( \Delta KE \). \( W_\text{total} = \Delta KE + \Delta U = -762.3 + 521.8 = -240.5 \, \text{J} \).
05

Determine Work Done by Friction

The work done by friction \( W_\text{friction} \) is the difference between total work and the change in potential energy: \( W_\text{friction} = W_\text{total} - \Delta U \). Substituting the values, \( W_\text{friction} = -762.3 - 521.8 \approx -1284.1 \, \text{J} \).
06

Calculate Kinetic Frictional Force

The magnitude of the kinetic frictional force \( f_k \) can be calculated using \( W_\text{friction} = f_k \cdot \text{distance} \cdot \cos(180^{\circ}) \). Given \( W_\text{friction} = -1284.1 \, \text{J} \) and distance = 1.9 m, \( f_k = \frac{-1284.1}{1.9 \, \cos(180^{\circ})} = \frac{1284.1}{1.9} \approx 676.9 \, \text{N} \).

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Key Concepts

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

Kinetic Frictional Force
When a skier moves uphill, there is a force opposing this motion called the kinetic frictional force. This frictional force happens because the skis slide across the snow, and it acts in the opposite direction to the skier's movement. It's part of why the skier's speed decreases as they ascend the hill.
To figure out the work done by this force, we use the formula:
  • \[ W_{\text{friction}} = f_k \cdot d \cdot \cos(\theta) \]
This equation means you multiply the frictional force \( f_k \) by the distance covered \( d \) and the cosine of the angle between the force direction and the displacement. Since the frictional force acts directly opposite the motion, the angle here is \(180^{\circ}\), making \( \cos(180^{\circ}) = -1 \).
  • Kinetic friction is crucial in energy computations and affects the work-energy equation.
Understanding this concept helps explain why the skier's energy turns into heat due to friction.
Gravitational Potential Energy
As the skier moves uphill, gravity also plays a key role by influencing their potential energy. Gravitational potential energy (GPE) is the energy stored because of an object's position in a gravitational field. For the skier, climbing the slope involves gaining height, which increases their potential energy.
The potential energy change \( \Delta U \) reflects the energy transferred to an object as it moves upward:
  • \[ \Delta U = mgh \]
Where \(m\) is the mass, \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)), and \(h\) is the change in height. Here, \(h\) is determined by the vertical component of the skier's path, calculated using \( 1.9 \times \sin(25^{\circ}) \). This increase in GPE comes into play by reducing the skier's kinetic energy as they climb up.
Kinetic Energy Change
Kinetic energy (KE) is the energy an object possesses due to its motion. As the skier moves up the slope, they start with an initial kinetic energy due to their initial speed (\(v_i = 6.6 \, \text{m/s} \)). As they move upward, the frictional force and the incline slow them down, leading to a decrease in kinetic energy.
  • \[ \Delta KE = \frac{1}{2}m(v_f^2 - v_i^2) \]
Where \(v_f\) is the final speed, and \(m\) is the mass of the skier. This change in kinetic energy shows how the energy is transferred or transformed during the skier's motion up the slope. Precisely, it decreases by about \(-762.3 \, \text{J}\) as calculated.
  • A decrease indicates that the skier is losing energy to both friction and potential energy gain.
This concept ties back to the work-energy principle, showing how energy conservation manages the skier's energy budget during the ascent.

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Most popular questions from this chapter

A 0.60 -kg basketball is dropped out of a window that is \(6.1 \mathrm{~m}\) above the ground. The ball is caught by a person whose hands are \(1.5 \mathrm{~m}\) above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) How is the change \(\left(\mathrm{PE}_{\mathrm{f}} \mathrm{PE}_{0}\right)\) in the ball's gravitational potential energy related to the work done by its weight?

A \(1200-\mathrm{kg}\) car is being driven up a \(5.0^{\circ}\) hill. The frictional force is directed opposite to the motion of the car and has a magnitude of \(f=524 \mathrm{~N}\). A force \(\overrightarrow{\mathrm{F}}\) is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight \(\overrightarrow{\mathrm{W}}\) and the normal force \(\overrightarrow{\mathrm{F}}_{\mathrm{N}}\) directed perpendicular to the road surface. The length of the road up the hill is \(290 \mathrm{~m}\). What should be the magnitude of \(\overrightarrow{\mathbf{F}}\), so that the net work done by all the forces acting on the car is \(+150 \mathrm{~kJ}\) ?

When an \(81.0\) - \(\mathrm{kg}\) adult uses a spiral staircase to climb to the second floor of his house, his gravitational potential energy increases by \(2.0 \times 10^{3} \mathrm{~J}\). By how much does the potential energy of an \(18.0\) - \(\mathrm{kg}\) child increase when the child climbs a normal staircase to the second floor?

A \(2.40 \times 10^{2}-N\) force is pulling an \(85.0\) -kg refrigerator across a horizontal surface. The force acts at an angle of \(20.0^{\circ}\) above the surface. The coefficient of kinetic friction is \(0.200\), and the refrigerator moves a distance of \(8.00 \mathrm{~m}\). Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

During a tug-of-war, team A pulls on team B by applying a force of \(1100 \mathrm{~N}\) to the rope between them. How much work does team A do if they pull team B toward them a distance of \(2.0 \mathrm{~m} ?\)

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