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A 5.00-kg package slides 2.80 m down a long ramp that is inclined at 24.0 below the horizontal. The coefficient of kinetic friction between the package and the ramp is μk = 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 m/s at the top of the ramp, what is its speed after it has slid 2.80 m down the ramp?

Short Answer

Expert verified
(a) -38.92 J, (b) 55.92 J, (c) 0 J, (d) 17.0 J, (e) 3.41 m/s.

Step by step solution

01

Calculate the Force of Friction

First, determine the force of friction. The force of friction can be calculated using the formula fk=μkN, where N is the normal force. On an inclined plane, N=mgcosθ, where m=5.00 kg, g=9.81 m/s2, and θ=24.0. Calculate: N=5.009.81cos(24.0)44.85 Nfk=0.310×44.8513.90 N
02

Calculate Work Done by Friction

The work done by friction is given by Wf=fkdcos(180), where d=2.80 m is the distance. Wf=13.90×2.80×(1)Wf=38.92 J
03

Calculate Work Done by Gravity

The work done by gravity is given by Wg=mgdsinθ. Use the values m=5.00 kg,g=9.81 m/s2,θ=24.0,d=2.80 m.Wg=5.00×9.81×2.80×sin(24.0)55.92 J
04

Calculate Work Done by Normal Force

The work done by the normal force is zero because the normal force acts perpendicular to the direction of motion. Hence,Wn=0 J
05

Calculate Total Work Done

The total work done on the package is the sum of the work done by friction, gravity, and the normal force.Wtotal=Wf+Wg+WnWtotal=38.92+55.92+0=17.0 J
06

Calculate Final Speed

Use the work-energy principle to find the final speed of the package. The net work done on the package is equal to the change in kinetic energy.Wtotal=12mvf212mvi2Rearrange to find vf:17.0=125.00vf2125.002.202Solve for vf:17.0=125.00(vf24.84)17.0=2.50(vf24.84)6.8=vf24.84vf2=11.64vf=11.643.41 m/s

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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 force that opposes the motion of objects sliding against each other. It plays a crucial role in slowing things down. When an object moves across a surface, kinetic friction acts in the opposite direction of the object's movement.
In our exercise, the coefficient of kinetic friction (μk=0.310) tells us how "sticky" the two surfaces are. The normal force (N) presses the object against the surface. It depends on the object's weight and the angle of the slope.To calculate the force of kinetic friction, we use the equation:
  • fk=μk×N
This frictional force results in energy loss, affecting how objects move over time.
In scenarios like our inclined plane, understanding kinetic friction is essential to predict the motion and final speed of the object.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, and it's a simple machine used to make lifting objects easier. In physics problems, an inclined plane helps demonstrate how gravity, friction, and other forces interact.
When an object is on an inclined plane, the gravitational force can be split into two components:
  • One parallel to the plane, causing the object to slide down.
  • Another perpendicular to the plane, acting as the normal force.
For our 24.0-degree incline, the parallel component is calculated using mgsin(θ) and the perpendicular using mgcos(θ).These components allow us to determine how the object moves down the plane and how friction affects that motion. Analyzing such setups helps us study complex real-world interactions in a simpler manner.
Net Work
In physics, work is done when a force moves an object over a distance. Net work is the total work done by all forces on an object. It's vital in understanding how energy transfers between forms and how it affects an object's motion.
When calculating net work, consider contributions from all forces:
  • Work done by gravity.
  • Work done by friction (often negative since it opposes motion).
  • Work by normal force (usually zero as it's perpendicular to movement).
In our scenario, the net work done is:Wtotal=Wf+Wg+WnThis net work translates into a change in kinetic energy, as described by the work-energy principle. This principle is key to predicting how the object's speed changes after moving a certain distance. By adding or subtracting energy through work, we can determine the final velocity of the package sliding down the ramp.

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Most popular questions from this chapter

It is 5.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h (which uses up energy at the rate of 700 W), or you could walk it leisurely at 3.0 km/h (which uses energy at 290 W). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why does the more intense exercise burn up less energy than the less intense exercise?

(a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant k, what is the force constant of each half, in terms of k? (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal?) (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of k?

Three identical 8.50-kg masses are hung by three identical springs (Fig. E6.35). Each spring has a force constant of 7.80 kN/m and was 12.0 cm long before any masses were attached to it. (a) Draw a free-body diagram of each mass. (b) How long is each spring when hanging as shown? (Hint: First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)

A soccer ball with mass 0.420 kg is initially moving with speed 2.00 m/s. A soccer player kicks the ball, exerting a constant force of magnitude 40.0 N in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 m/s?

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Fig. P6.71). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is 2.80 m/s. (a) What is the tension in the cord in the original situation, when the block has speed υ=0.70 m/s? (b) What is the tension in the cord in the final situation, when the block has speed υ=2.80 m/s? (c) How much work was done by the person who pulled on the cord?

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