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Chapter 5: Problem 5

Starting from rest, a \(5.00-\mathrm{kg}\) block slides \(2.50 \mathrm{~m}\) down a rough \(30.0^{\circ}\) incline. The coefficient of kinetic friction between the block and the incline is \(\mu_{k}=0.496 .\) Determine (a) the work done by the force of gravity, (b) the work done by the friction force between block and incline, and (c) the work done by the normal force. (d) Qualitatively, how would the answers change if a shorter ramp at a steeper angle were used to span the same vertical height?

Short Answer

Expert verified
The work done by the force of gravity is \( W_g = m \cdot g \cdot h \cdot \sin(30) \) and work done by the friction force is \( W_f = - \mu \cdot F_n \cdot h \). The work done by the normal force is 0. If a shorter ramp at steeper angle were used to span the same vertical height, less work would be done by gravity because the distance the block would be slidden would reduce, however, more work would be done by friction because the normal force would increase, leading to increased friction.

Step by step solution

01

Work Done by Gravity

The work done by the force of gravity is equal to the force of gravity times the distance the object moves in the direction of the force. The force of gravity is given by \( F = m \cdot g \), where \( m = 5.0 \, \text{kg} \) is the mass and \( g \approx 9.81 \, \text{m/s}^2 \) is the gravitational acceleration. The block slides a distance \( h = 2.5 \, \text{m} \) down the incline which can be broken into vertical \( (h_y) \) and horizontal \( (h_x) \) components by using the formula \( h = h_y \cdot \sin(30) + h_x \cdot \cos(30) \). The work done by gravity is thus calculated as \( W_g = m \cdot g \cdot h \cdot \sin(30) \).
02

Work Done by Friction

The work done by the friction force is equal to the friction force times the distance the object moves in the direction of the force. The friction force is given by \( F = \mu \cdot F_n \), where \( \mu = 0.496 \) is the coefficient of kinetic friction and \( F_n = m \cdot g \cdot \cos(30) \) is the normal force. The block slides a distance \( h = 2.5 \, \text{m} \) down the incline. The work done by friction is thus calculated as \( W_f = \mu \cdot F_n \cdot h \) and with the minus sign, since it is acting against the motion.
03

Work Done by the Normal Force

The work done by the normal force is zero because the normal force always acts perpendicular to the direction of the motion.
04

Explaining the qualitative difference

If a shorter ramp at a steeper angle were used to span the same vertical height, the distance the block moves due to gravity would be shorter. This would mean that less work would be done by gravity. However, because the angle of the incline is steeper, the normal force and consequently the frictional force would be larger. This would mean that more work would be done by friction. The work done by the normal force would remain zero.

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Key Concepts

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

Kinetic Friction
Kinetic friction occurs when two surfaces are moving relative to each other and oppose the motion. In physics, it's an important factor when analyzing the movement of objects, especially on inclined planes. The kinetic frictional force is calculated using the formula:

\( F_f = \boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol\mu_k} \times F_n \)

Here, \( \boldsymbol{\mu_k} \) is the coefficient of kinetic friction and \( F_n \) is the normal force acting on the object. The kinetic frictional force always acts in the direction opposite to the object鈥檚 motion and is a factor that resists the object's sliding.
Normal Force
The normal force is the force that a surface exerts on an object to support the weight of the object. It always acts perpendicular to the surface. For objects on a flat surface, the normal force is typically equal to the gravitational force. However, for objects on an inclined plane, the normal force is less, because it is equal to the component of the gravitational force that is perpendicular to the plane.

On an incline, the normal force is given by: \( F_n = m \times g \times \boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol{\boldsymbol\cos}(\theta) \)

where \( m \) is the mass, \( g \) is the gravitational acceleration, and \( \theta \) is the angle of the incline.
Gravitational Force
Gravitational force is a fundamental interaction that attracts two bodies with mass. In the context of our problem, it's what causes the block to slide down the incline. It is calculated simply by the product of the mass and the gravitational acceleration: \( F_g = m \times g \)

Here, \( m \) is the mass of the block and \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{m/s}^2 \). The component of this force that does the work as the block slides down the incline is parallel to the plane, and it is calculated by multiplying the force of gravity by the sine of the angle of the incline.
Inclined Plane Physics
When dealing with inclined planes, the motion and forces involved become more complex due to the angle of the incline. It affects both the gravitational force acting on the object and the magnitude of the normal force. For an object sliding down an incline without acceleration, the net force along the incline is zero, which means the friction force is equal in magnitude to the component of the gravitational force along the incline.

In the case of our block, the inclined plane alters how the gravitational force is applied, which in turn changes the normal force and frictional force. As the steepness of the incline increases, the component of the gravity force parallel to the incline increases as well, thus increasing the frictional force if the coefficient of kinetic friction remains the same.
Coefficient of Kinetic Friction
The coefficient of kinetic friction, symbolized as \( \boldsymbol{\mu_k} \), is a dimensionless value that represents the ratio of the force of kinetic friction between two surfaces to the normal force pressing them together. This value changes based on the materials that are in contact and how polished or rough their surfaces are. High values of \( \boldsymbol{\mu_k} \) indicate that the surfaces have a stronger resistance to sliding past each other.

Working with the coefficient of kinetic friction helps us to understand and calculate the forces at play when an object is moving across another surface. In our exercise, the given coefficient helps us to determine how much the friction force opposes the block's movement down the incline, and thus, influences the work done by this frictional force.

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Most popular questions from this chapter

A 2 \(300-\mathrm{kg}\) pile driver is used to drive a steel beam into the ground. The pile driver falls \(7.50 \mathrm{~m}\) before coming into contact with the top of the beam, and it drives the beam \(18.0 \mathrm{~cm}\) farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.

In bicycling for aerobic exercise, a woman wants her heart rate to be between 196 and 166 beats per minute. Assume that her heart rate is directly proportional to her mechanical power output. Ignore all forces on the womanplus-bicycle system, except for static friction forward on the drive wheel of the bicycle and an air resistance force proportional to the square of the bicycler's speed. When her speed is \(22.0 \mathrm{~km} / \mathrm{h}\), her heart rate is \(90.0\) beats per minute. In what range should her speed be so that her heart rate will be in the range she wants?

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?

A ski jumper starts from rest \(50.0 \mathrm{~m}\) above the ground on a frictionless track and flies off the uack at an angle of \(45.0^{\circ}\) above the horizontal and at a height of \(10.0 \mathrm{~m}\) above the level ground. Neglect air resistance. (a) What is her speed when she leaves the track? (b) What is the maximum altitude she attains after leaving the track? (c) Where does she land relative to the end of the track?

A horizontal spring attached to a wall has a force constant of \(850 \mathrm{~N} / \mathrm{m} . \mathrm{A}\) block of mass \(1.00 \mathrm{~kg}\) is attached to the spring and oscillates freely on a horizontal, frictionless surface as in Active Figure \(5.20 .\) The initial goal of this problem is to find the velocity at the equilibrium point after the block is released. (a) What objects constitute the system, and through what forces do they interact? (b) What are the two points of interest? (c) Find the energy stored in the spring when the mass is stretched \(6.00 \mathrm{~cm}\) from equilibrium and again when the mass passes through cquilibrium after being released from rest. (d) Write the conservation of energy equation for this situation and solve it for the speed of the mass as it passes equilibrium. Substitute to obtain a numerical value. (e) What is the speed at the halfway point? Why isn't it half the speed at equilibrium?
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