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A truck travels uphill with constant velocity on a highway with a \(7.0^{\circ}\) slope. A 50 -kg package sits on the floor of the back of the truck and does not slide, due to a static frictional force. During an interval in which the truck travels \(340 \mathrm{~m}\), what is the net work done on the package? What is the work done on the package by the force of gravity, the normal force, and the friction force?

Short Answer

Expert verified
The net work done on the package is \(-1.946 \times 10^4 \, J\). The work done by gravity is \(-1.946 \times 10^4 \, J\), by the normal force is 0 J, and by the friction force is also 0 J.

Step by step solution

01

Identify Given Information

By considering the problem, one can identify the following given information: the mass of the package (\(m = 50 \, kg\)), the distance the truck travels (\(d = 340 \, m\)), and the angle of the slope (\(\theta = 7.0^{\circ}\)).
02

Calculate Work Done by Gravity

The work done on the package by gravity is the gravitational force times the displacement in the direction of the force. The displacement in the direction of gravity is \(d \sin(\theta)\), and the gravitational force is \(mg\). So the work done by gravity (\(W_g\)) is \(mgd \sin(\theta)\). Converting the angle into radians (as sin function works with radians) and substituting the given values, we obtain \(W_g = 50 \, kg \cdot 9.8 \, m/s^2 \cdot 340 \, m \cdot \sin(7.0^{\circ}) = -1.946 \times 10^4 \, J\). The minus sign indicates that the work is done against the direction of displacement.
03

Calculate Work Done by Normal Force

The Normal force acts perpendicular to the displacement of the package. Therefore, the angle between the displacement and the Normal force is 90 degrees. The cosine of 90 degrees is zero, so the work done by the Normal force \(W_N\) is zero.
04

Calculate Work Done by Frictional Force

In this problem, given the package does not slide, the frictional force is static and acts perpendicular to the direction of displacement. Therefore, similar to the Normal force, the work done by the frictional force (\(W_f\)) is also zero.
05

Calculate the Net Work Done

The net work done on the package is the sum of the works done by gravity, the Normal force, and the frictional force. This is given by \(W_g+ W_N+ W_f = -1.946 \times 10^4 \, J+ 0+ 0 = -1.946 \times 10^4 \, J\).

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Key Concepts

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

Static Frictional Force
Understanding static frictional force is fundamental when analyzing motion scenarios where an object remains stationary relative to a moving surface, like the package inside the uphill-travelling truck. Static friction is the force that counteracts external forces and prevents two surfaces from sliding past each other. It is not a constant force but rather adjusts up to a maximum value based on the nature of the two surfaces in contact and the normal force pressing them together.

Here's a handy metaphor: Picture static friction as a superhero who holds an object in place against a villain—the pulling force. The stronger the villain pulls (up to a point), the harder our superhero pushes back, keeping the object safe—stationary. In our truck example, the static frictional force exactly matches the component of gravitational force acting downhill, because no slipping occurs, and therefore, no work is done by static friction, since there is no displacement along the direction of friction.

Key Takeaway

Static friction is a self-adjusting force that acts to prevent motion and will exert just enough force to maintain equilibrium until the threshold, known as the coefficient of static friction, is surpassed. If an object does not move relative to the contact surface, no work is done by static friction, as there's no movement to use force over a distance.
Gravitational Force
Gravitational force, a fundamental force in the universe, pulls objects toward one another, with its most familiar effect being objects falling towards Earth. This force is the reason why the package in the truck scenario has weight and why work can be done on it as the truck moves uphill.

The gravitational force exerted on an object near Earth is the product of the object's mass and the acceleration due to gravity (\(9.8 \text{ m/s}^2\)). In our example, it's the component of the weight of the package acting in the direction of the truck's displacement up the slope that's pertinent. The formula for the work done by gravity is the force times displacement in the direction of the force (\(W_g = mgd \text{ sin}(\theta)\)), demonstrating that gravity does work on the package as the truck climbs the hill, even though the package itself is not sliding within the truck.

Core Understanding

Gravitational force is a constant presence and its effects on work become apparent when objects are moved vertically or along a slope. Work is a measure of energy transfer, and when gravity does work, energy is transferred from one type to another, for instance, from potential to kinetic energy.
Normal Force
The concept of the normal force is crucial in understanding contact forces, which arise whenever two surfaces interact. The normal force is always perpendicular to the surfaces in contact. It's a reactive force, which means it responds to other forces to keep solid objects from passing through each other.

In the truck and package scenario, the normal force is exerted by the truck bed on the package, pushing it upward and perpendicular to the bed. This force counteracts the perpendicular component of the package's weight. Since the normal force and the displacement are always perpendicular, no work is done by the normal force. Remember, work is only accomplished when a force causes displacement in the direction of the force applied. This is why, despite the immense power normal force can provide, it will not account for work in scenarios where there is no displacement in its direction.

Essential Insight

Normal force plays a significant role in balancing forces in vertical or inclined scenarios and contributes to static frictional force but does not directly contribute to work unless there's movement in the direction it acts. This reinforces the fact that the presence of force does not always equate to the performance of work.

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Most popular questions from this chapter

When an automobile moves with constant speed down a highway, most of the power developed by the engine is used to compensate for the mechanical energy loss due to frictional forces exerted on the car by the air and the road. If the power developed by an engine is \(175 \mathrm{hp}\), estimate the total frictional force acting on the car when it is moving at a speed of \(29 \mathrm{~m} / \mathrm{s}\). One horsepower equals \(746 \mathrm{~W}\).

A daredevil wishes to bungee-jump from a hot-air balloon \(65.0 \mathrm{~m}\) above a carnival midway (Fig. P5.83). He will use a piece of uniform elastic cord tied to a harness around his body to stop his fall at a point \(10.0 \mathrm{~m}\) above the ground. Model his body as a particle and the cord as having negligible mass and a tension force described by Hooke's force law. In a preliminary test, hanging at rest from a \(5.00-\mathrm{m}\) length of the cord, the jumper finds that his body weight stretches it by \(1.50 \mathrm{~m} .\) He will drop from rest at the point where the top end of a longer section of the cord is attached to the stationary balloon. (a) What length of cord should he use? (b) What maximum acceleration will he experience?

A \(1.50 \times 10^{3}\) -kg car starts from rest and accelerates uniformly to \(18.0 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\). Assume that air resistance remains constant at \(400 \mathrm{~N}\) during this time. Find (a) the average power developed by the engine and (b) the instantaneous power output of the engine at \(t=12.0 \mathrm{~s}\), just before the car stops accelerating.

A shopper in a supermarket pushes a cart with a force of \(95 \mathrm{~N}\) directed at an angle of \(25^{\circ}\) below the horizontal. The force is just sufficient to overcome various frictional forces, so the cart moves at constant speed. (a) Find the work done by the shopper as she moves down a \(50.0-\mathrm{m}\) length aisle, (b) What is the net work done on the cart? Why? (c) The shopper goes down the next aisle, pushing horizontally and maintaining the same speed as before. If the work done by frictional forces doesn't change, would the shopper's applied force be larger, smaller, or the same? What about the work done on the cart by the shopper?

The masses of the javelin, discus, and shot are \(0.80 \mathrm{~kg}, 2.0 \mathrm{~kg}_{1}\) and \(7.2 \mathrm{~kg}\), respectively, and record throws in the corresponding track events are about \(98 \mathrm{~m}, 74 \mathrm{~m}_{1}\) and \(23 \mathrm{~m}\), respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws, and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of \(2.0 \mathrm{~m} .\) (c) Do your results suggest that air resistance is an important factor?

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