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Chapter 5: Problem 50

\(\bullet$$\bullet\) You are working for a shipping company. Your job is to stand at the bottom of an 8.0 -m-long ramp that is inclined at \(37^{\circ}\) above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is \(\mu_{k}=0.30\) (a) What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp? (b) Your coworker is supposed to grab the packages as they arrive at the top of the ramp, but she misses one and it slides back down. What is its speed when it returns to you?

Short Answer

Expert verified
(a) Initial speed needed: 5.96 m/s. (b) Speed at bottom: 4.01 m/s.

Step by step solution

01

Understand the Problem

We have a ramp inclined at \( 37^\circ \) of length 8 meters. We need to calculate the initial velocity required to push a package up the ramp such that it comes to rest at the top. Additionally, we want to find the speed of the package as it slides back down to the bottom.
02

Calculate Forces on the Package

Identify the forces on the package: gravitational force, normal force, and frictional force. Gravitational force acts downwards parallel to the ramp as \( mg \sin(37^\circ) \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (\( 9.8\, \text{m/s}^2 \)). The normal force becomes \( mg \cos(37^\circ) \). Kinetic friction is \( F_k = \mu_k N = \mu_k mg \cos(37^\circ) \).
03

Apply Energy Conservation for the Ramp Upward Movement

As the package travels up the ramp, its initial kinetic energy is converted to overcome gravitational potential energy and work against friction. The equation is:\[ \frac{1}{2} m v_i^2 = mg h + F_k d \]where \( h = d \sin(37^\circ) \) and \( d = 8 \). Calculate \( v_i \) using the equation.
04

Simplify the Equation and Solve for Initial Velocity

Substitute the values into the equation: \[ \frac{1}{2} m v_i^2 = mg (8 \sin(37^\circ)) + 0.30 mg (8 \cos(37^\circ)) \]Cancel out \( m \) from both sides:\[ v_i^2 = 2g (8 \sin(37^\circ) + 0.30 \cdot 8 \cos(37^\circ)) \]Calculate \( v_i \) using the above expression.
05

Calculate Speed When Returning Downward

Assuming no energy is lost other than due to friction, use conservation of energy principles again. The energy at the top (all potential) is \[ mg d \sin(37^\circ) \]This potential energy is converted back to kinetic energy and work done by friction on return. Solve for \( v_f \):\[ \frac{1}{2} m v_f^2 = mg h - F_k d \]or\[ v_f^2 = 2g(8 \sin(37^\circ) - 0.30 \cdot 8 \cos(37^\circ)) \]Calculate \( v_f \).
06

Compute Final Numerical Results

Calculate numerical values using known constants. The gravitational constant \( g = 9.8 \text{ m/s}^2 \), \( \sin(37^\circ) \approx 0.6018 \), \( \cos(37^\circ) \approx 0.7986 \), and \( d = 8 \text{ m} \). Substitute these into the expressions for \( v_i \) and \( v_f \) to find final numerical results.

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Key Concepts

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

Energy Conservation
Energy conservation is a fundamental concept in physics that helps us understand how energy is transferred and transformed. In many physics problems, it is advantageous to think of energy in terms of how it changes. Understanding how energy conservation applies to this scenario involves recognizing different forms of energy:
  • Kinetic Energy: This is the energy associated with motion. At the bottom of the ramp, the package has kinetic energy due to its speed. Kinetic energy is given by the equation \[ KE = \frac{1}{2} mv^2 \]
  • Potential Energy: As the package moves up the ramp, it gains gravitational potential energy, which depends on its height and is expressed as \[ PE = mgh \]
  • Work Done by Friction: The kinetic friction between the ramp and the package does work on the package, removing energy from the system as the package moves.
The principle of energy conservation states that the initial energy (in the form of kinetic energy) is transformed into potential energy at the top of the ramp, minus the energy lost to friction. This helps us calculate the unknown velocity needed to start.
Inclined Plane
An inclined plane is a flat surface tilted at an angle, known in physics as a plane. It allows us to tap into the components of physical forces acting on an object. On an inclined plane:
  • Gravity: The weight of an object can be broken into components: parallel and perpendicular to the surface of the plane. The component parallel to the plane accelerates the object down the slope and is given by \[ mg \sin(\theta) \]
  • Normal Force: This force acts perpendicular to the plane's surface, providing an instance that counteracts the parallel component of gravity. Calculated as \[ mg \cos(\theta) \]
Understanding these forces helps find how objects move on an inclined plane. It aids in solving for parameters like initial velocity and time. The critical issue here is considering how these forces interact with kinetic friction to understand the net effect on the motion of the package.
Kinetic Friction
Kinetic friction occurs between two surfaces in relative motion. It plays a vital role in determining an object's motion on an inclined plane. This type of friction is crucial in energy conservation equations and affects how much initial speed is needed. The force of kinetic friction is determined by:
  • Coefficient of Kinetic Friction: Represented by \( \mu_k \), this dimensionless number indicates how much friction a material exerts when moving over another. For the ramp in the problem, \( \mu_k = 0.30 \).
  • Normal Force: The upward force exerted by the plane鈥檚 surface. The kinetic frictional force \( F_k \) is proportional to the normal force and given by \[ F_k = \mu_k N \]
Kinetic friction鈥檚 role is to oppose motion, converting mechanical energy into thermal energy and thus reducing the system's mechanical energy. Understanding kinetic friction is essential to calculate the initial push speed needed and the speed at which a package returns.
Newton's Laws of Motion
Newton's laws of motion form the foundation for our understanding of how objects move. In our ramp problem, applying Newton's laws helps determine forces and predict the movement of the package.
  • First Law: A body at rest will remain at rest, and a body in motion will remain in motion unless acted upon by a force. When the package is stationary, it requires an external force to start moving.
  • Second Law: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the force. Calculated as \[ F = ma \], where \( F \) is the force applied, \( m \) is the mass, and \( a \) is the acceleration. Newton's second law helps us understand how the forces combine to result in acceleration down the ramp.
  • Third Law: For every action, there is an equal and opposite reaction. This is evident in the normal force and gravitational force interplay.
By comprehending these laws, we predict how the initial velocity and friction interact to slow the package down as it moves up the ramp and accelerates it back when descending.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Air-bag safety. According to safety standards for air bags, the maximum acceleration during a car crash should not exceed 60\(g\) and should last for no more than 36 ms. (a) In such a case, what force does the air bag exert on a 75 kg person? Start with a free-body diagram. (b) Express the force in part (a) in terms of the person's weight.

\(\bullet$$\bullet\) Tension in a muscle. Muscles are attached to bones by means of tendons. The maximum force that a muscle can exert is directly proportional to its cross-sectional area \(A\) at the widest point. We can express this relationship mathematically as \(F_{\text { max }}=\sigma A,\) where \(\sigma(\) sigma) is a proportionality constant. Surprisingly, \(\sigma\) is about the same for the muscles of all animals. and has the numerical value of \(3.0 \times 10^{5}\) in SI units. (a) What are the SI units of \(\sigma\) in terms of newtons and meters and also in terms of the fundamental quantities \((\mathrm{kg}, \mathrm{m}, \mathrm{s})\) (b) In one set of experiments, the average maximum force that the gastrocne- mius muscle in the back of the lower leg could exert was meas- ured to be 755 \(\mathrm{N}\) for healthy males in their midtwenties. What does this result tell us was the average cross-sectional area, in \(\mathrm{cm}^{2},\) of that muscle for the people in the study?

\(\bullet$$\bullet\) You've attached a bungee cord to a wagon and are using it to pull your little sister while you take her for a jaunt. The bungee's unstretched length is \(1.3 \mathrm{m},\) and you happen to know that your little sister weighs 220 \(\mathrm{N}\) and the wagon weighs 75 \(\mathrm{N}\) . Crossing a street, you accelerate from rest to your normal walk- ing speed of 1.5 \(\mathrm{m} / \mathrm{s}\) in 2.0 \(\mathrm{s}\) , and you notice that while you're accelerating, the bungee's length increases to about 2.0 \(\mathrm{m}\) . What's the force constant of the bungee cord, assuming it obeys Hooke's law?

\(\bullet\) An unstretched spring is 12.00 \(\mathrm{cm}\) long. When you hang an 875 g weight from it, it stretches to a length of 14.40 \(\mathrm{cm}\) . (a) What is the force constant (in \(\mathrm{N} / \mathrm{m}\) ) of this spring? (b) What total mass must you hang from the spring to stretch it to a total length of 17.72 \(\mathrm{cm} ?\)

\(\bullet$$\bullet\) Force during a jump. An average person can reach a maximum height of about 60 \(\mathrm{cm}\) when jumping straight up from a crouched position. During the jump itself, the person's body from the knees up typically rises a distance of around 50 \(\mathrm{cm} .\) To keep the calculations simple and yet get a reason- able result, assume that the entire body rises this much during the jump. (a) With what initial speed does the person leave the ground to reach a height of 60 \(\mathrm{cm} ?\) (b) Make a free-body diagram of the person during the jump. (c) In terms of this jumper's weight \(W,\) what force does the ground exert on him or her during the jump?
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