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A 5.00-kg package slides 2.80 m down a long ramp that is inclined at 24.0\(^\circ\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_k\) \(=\) 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 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 \( f_k = \mu_k \cdot N \), where \( N \) is the normal force. On an inclined plane, \( N = mg\cos\theta \), where \( m = 5.00 \text{ kg} \), \( g = 9.81 \text{ m/s}^2 \), and \( \theta = 24.0^\circ \). Calculate: \[ N = 5.00 \cdot 9.81 \cdot \cos(24.0^\circ) \approx 44.85 \text{ N} \]\[ f_k = 0.310 \times 44.85 \approx 13.90 \text{ N} \]
02

Calculate Work Done by Friction

The work done by friction is given by \( W_f = f_k \cdot d \cdot \cos(180^\circ) \), where \( d = 2.80 \text{ m} \) is the distance. \[ W_f = 13.90 \times 2.80 \times (-1) \]\[ W_f = -38.92 \text{ J} \]
03

Calculate Work Done by Gravity

The work done by gravity is given by \( W_g = mgd\sin\theta \). Use the values \( m = 5.00 \text{ kg}, g = 9.81 \text{ m/s}^2, \theta = 24.0^\circ, d = 2.80 \text{ m} \).\[ W_g = 5.00 \times 9.81 \times 2.80 \times \sin(24.0^\circ) \approx 55.92 \text{ J} \]
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,\[ W_n = 0 \text{ J} \]
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.\[ W_{total} = W_f + W_g + W_n \]\[ W_{total} = -38.92 + 55.92 + 0 = 17.0 \text{ J} \]
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.\[ W_{total} = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2 \]Rearrange to find \( v_f \):\[ 17.0 = \frac{1}{2} \cdot 5.00 \cdot v_f^2 - \frac{1}{2} \cdot 5.00 \cdot 2.20^2 \]Solve for \( v_f \):\[ 17.0 = \frac{1}{2} \cdot 5.00 \cdot (v_f^2 - 4.84) \]\[ 17.0 = 2.50(v_f^2 - 4.84) \]\[ 6.8 = v_f^2 - 4.84 \]\[ v_f^2 = 11.64 \]\[ v_f = \sqrt{11.64} \approx 3.41 \text{ m/s} \]

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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 (\( \mu_k = 0.310 \)) 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:
  • \( f_k = \mu_k \times N \)
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 \( mg\sin(\theta) \) and the perpendicular using \( mg\cos(\theta) \).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:\( W_{total} = W_f + W_g + W_n \)This 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

An airplane in flight is subject to an air resistance force proportional to the square of its speed v. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (\(\textbf{Fig. P6.94}\)). The upward force is the lift force that keeps the airplane aloft, and the backward force is called \(induced \, drag\). At flying speeds, induced drag is inversely proportional to \(v^2\), so the total air resistance force can be expressed by \(F_air = \alpha v^{2} + \beta /v{^2}\), where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna 150, a small single-engine airplane, \(\alpha = 0.30 \, \mathrm{N} \cdot \mathrm{s^{2}/m^{2}}\) and \(\beta = 3.5 \times 10^5 \, \mathrm{N} \cdot \mathrm{m^2/s^2}\). In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in km/h) at which this airplane will have the maximum \(range\) (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in km/h) for which the airplane will have the maximum \(endurance\)(that is, remain in the air the longest time).

A physics student spends part of her day walking between classes or for recreation, during which time she expends energy at an average rate of 280 W. The remainder of the day she is sitting in class, studying, or resting; during these activities, she expends energy at an average rate of 100 W. If she expends a total of 1.1 \(\times\) 10\(^7\) J of energy in a 24-hour day, how much of the day did she spend walking?

A 4.80-kg watermelon is dropped from rest from the roof of an 18.0-m-tall building and feels no appreciable air resistance. (a) Calculate the work done by gravity on the watermelon during its displacement from the roof to the ground. (b) Just before it strikes the ground, what is the watermelon's (i) kinetic energy and (ii) speed? (c) Which of the answers in parts (a) and (b) would be \(different\) if there were appreciable air resistance?

A spring of force constant 300.0 N/m and unstretched length 0.240 m is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 N. How long will the spring now be, and how much work was required to stretch it that distance?

We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M\), equilibrium length \(L_0\), and force constant \(k\). The work done to stretch or compress the spring by a distance \(L\) is \\(\frac{1}{2}\\) \(kX^2\), where \(X = L - L_0\). Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v\). Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of \(M\) and \(v\). (\(Hint\): Divide the spring into pieces of length \(dl\); find the speed of each piece in terms of \(l\), \(v\), and \(L\); find the mass of each piece in terms of \(dl\), \(M\), and \(L\); and integrate from \(0\) to \(L\). The result is \(not\) \\(\frac{1}{2}\\) \(Mv^2\), since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 kg and force constant 3200 N/m is compressed 2.50 cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 0.053-kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

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