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A 5.00-kg package slides 1.50 \(\mathrm{m}\) down a long ramp that is inclined at \(12.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_{\mathrm{k}}=0.310\) . Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity, (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) at the top of the ramp, what is its speed after sliding 1.50 \(\mathrm{m}\) down the ramp?

Short Answer

Expert verified
(a) Work by friction: Approximately -20.5 J, (b) Gravity: Approximately 15.4 J, (c) Normal force: 0 J, (d) Total work: Approximately -5.1 J, (e) Final speed: Approximately 1.64 m/s.

Step by step solution

01

Determine the force of friction

First, calculate the normal force \( F_N \) which acts perpendicular to the ramp. For an object on an incline,\[F_N = mg \cos(\theta)\]where \( m = 5.00 \text{ kg} \) is the mass of the package, \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity, and \( \theta = 12.0^\circ \) is the angle of the incline. Substitute the values to compute \( F_N \).Next, determine the frictional force \( F_f \) using the coefficient of kinetic friction \( \mu_k = 0.310 \):\[F_f = \mu_k \times F_N = 0.310 \times (5.00 \times 9.81 \cos(12.0^\circ))\]
02

Calculate work done by friction

The work done by friction \( W_f \) can be calculated as:\[W_f = F_f \times d \times \cos(180^\circ)\]where \( d = 1.50 \text{ m} \) is the distance slid down the ramp and \( \cos(180^\circ) = -1 \) because friction acts opposite to the direction of movement. Calculate \( W_f \).
03

Calculate work done by gravity

The work done by gravity \( W_g \) is given by:\[W_g = mg \times d \times \sin(\theta)\]Calculate \( W_g \) using the known values for mass, gravitational acceleration, the angle of the incline, and the distance.
04

Determine work done by the normal force

The work done by the normal force on an object moving along an incline is zero because the normal force acts perpendicular to the direction of displacement:\[W_N = 0\]
05

Calculate total work done

The total work \( W_{total} \) is the sum of work done by friction, gravity, and the normal force:\[W_{total} = W_f + W_g + W_N\]Substitute the calculated values from previous steps to compute \( W_{total} \).
06

Use work-energy theorem to find final speed

Apply the work-energy principle:\[K_f = K_i + W_{total}\]where \( K_i \) is the initial kinetic energy:\[K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 5.00 \times (2.20)^2\]and \( K_f \) is the final kinetic energy:\[K_f = \frac{1}{2} m v_f^2\]Rearrange the equation to find the final speed \( v_f \):\[ v_f = \sqrt{v_i^2 + \frac{2 \times W_{total}}{m} } \]Substitute the known values to solve for \( v_f \).

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Key Concepts

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

Kinetic Friction
Kinetic friction is a type of frictional force that occurs when two surfaces slide against each other. In this situation, when a package is sliding down a ramp, the force of kinetic friction opposes the motion. It always acts in the opposite direction of the object's movement, causing it to slow down.
To calculate the force of kinetic friction, you need the coefficient of kinetic friction (\(\mu_k\)) and the normal force (\( F_N \)). The normal force is the support force exerted by a surface against an object resting on it and is perpendicular to the surface. For an inclined plane, it is given by:
  • \( F_N = m \cdot g \cdot \cos(\theta) \)
where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( \theta \) is the angle of the incline.
Once you have the normal force, you can calculate the force of kinetic friction as:
  • \( F_f = \mu_k \cdot F_N \)
This force is used in computing the work done by friction, given by the formula:
  • \( W_f = F_f \cdot d \cdot \cos(180^{\circ}) \)
where \( d \) is the distance traveled, and the angle is \( 180^{\circ} \) because friction opposes the motion.
Inclined Plane
An inclined plane is a flat surface that is tilted at an angle to the horizontal. It is used to demonstrate how forces and energy interact differently on a sloped surface compared to a flat one.
When an object moves along an inclined plane, the gravitational force acting on it can be split into two components:
  • Parallel to the inclined plane: \( F_{ ext{parallel}} = mg \sin(\theta) \)
  • Perpendicular to the inclined plane: \( F_{ ext{perpendicular}} = mg \cos(\theta) \)
The force parallel to the incline is what causes the object to slide down. This inclination changes how energy and forces such as gravity and friction affect the motion.
The gravitational force component parallel to the incline, combined with kinetic friction, determines the work done on the object as it moves. Incorporating their effect helps us understand how the inclined plane modifies the typical horizontal and vertical force interpretations, providing invaluable insights into physics concepts.
Work-Energy Theorem
The work-energy theorem is a fundamental concept in physics that connects the principles of work and energy. It states that the work done on an object is equal to the change in its kinetic energy.
In practical terms, this means:
  • If the total work done on an object is positive, its kinetic energy increases, and it speeds up.
  • If the total work done is negative, its kinetic energy decreases, and it slows down.
For the exercise involving the sliding package, we calculate the initial kinetic energy \( K_i \) using the velocity at the top of the ramp:
  • \( K_i = \frac{1}{2} m v_i^2 \)
Then, we find the total work done by adding up all the work done by the different forces:
  • \( W_{\text{total}} = W_f + W_g + W_N \)
Here, \( W_N \) is zero because the normal force does not work when acting perpendicular to displacement. Finally, we use these values to find the final speed at the bottom of the ramp using the work-energy theorem:
  • \( v_f = \sqrt{v_i^2 + \frac{2 \times W_{\text{total}}}{m} } \)
This seamless application of energy conservation and force analysis beautifully encapsulates the work-energy principle in dynamics.

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Most popular questions from this chapter

The total consumption of electrical energy in the United States is about 1.0 \(\times 10^{19} \mathrm{J}\) per year. (a) What is the average rate of electrical energy consumption in watts? (b) The population of the United States is about 300 million people. What is the average rate of electrical energy consumption per person? (c) The sun transfers energy to the earth by radiation at a rate of approximately 1.0 \(\mathrm{kW}\) per square meter of surface. If this energy could be collected and converted to electrical energy with 40\(\%\) efficiency, how great an area (in square kilometers) would be required to collect the electrical energy used in the United States?

To stretch a spring 3.00 \(\mathrm{cm}\) from its unstretched length, 12.0 \(\mathrm{J}\) of work must be done. (a) What is the force constant of this spring? (b) What magnitude force is needed to stretch the spring 3.00 \(\mathrm{cm}\) from its unstretched length? (c) How much work must be done to compress this spring 4.00 \(\mathrm{cm}\) from its unstretched length, and what force is needed to streteh it this distance?

An object is attracted toward the origin with a force given by \(F_{x}=-k / x^{2} .\) (Gravitational and electrical forces have this distance dependence.) (a) Calculate the work done by the force \(F_{x}\) when the object moves in the \(x\) -direction from \(x_{1}\) to \(x_{2}\) . If \(x_{2}>x_{1}\) , is the work done by \(F_{x}\) positive or negative? (b) The only other force acting on the object is a force that you exert with your hand to move the object slowly from \(x_{1}\) to \(x_{2}\) . How much work do you do? If \(x_{2}>x_{1},\) is the work you do positive or negative? (c) Explain the similarities and differences between your answers to parts (a) and (b).

You are a member of an Alpine Rescue Team. You must project a box of supphes up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathbf{k}}\) . Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha .\)

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