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A crate is being pushed across a rough floor surface. If no force is applied on the crate, the crate will slow down and come to a stop. If the crate of mass \(50 \mathrm{kg}\) moving at speed \(8 \mathrm{m} / \mathrm{s}\) comes to rest in 10 seconds, what is the rate at which the frictional force on the crate takes energy away from the crate?

Short Answer

Expert verified
The rate at which the frictional force takes energy away from the crate is \(160 \mathrm{W}\).

Step by step solution

01

Calculate the change in velocity

Given that the crate starts at a velocity of 8 m/s and comes to rest, the change in velocity is the final velocity minus the initial velocity: \(∆v = 0 - 8 \mathrm{m/s} = -8 \mathrm{m/s}\).
02

Calculate the total time taken by the crate to come to a stop

Given the crate comes to rest in 10 seconds, total time taken by the crate to come to a stop is: \(t = 10 \mathrm{s}\).
03

Calculate the acceleration due to friction

Acceleration is the change in velocity divided by the time. In this case, the acceleration is due to friction, so we have: \(a = \frac{∆v}{t} = \frac{-8 \mathrm{m/s}}{10 \mathrm{s}} = -0.8 \mathrm{m/s^2}\).
04

Calculate the frictional force

The frictional force can be calculated using the formula \(F = ma\), where \(m\) is the mass of the crate and \(a\) is the acceleration due to friction. \(F = (50 \mathrm{kg})(-0.8 \mathrm{m/s^2}) = -40 \mathrm{N}\). Since friction acts in the opposite direction, we can say the frictional force is 40 N.
05

Calculate the work done by friction

The work done by friction can be calculated using the formula \(W = Fd\), where \(F\) is the frictional force and \(d\) is the distance traveled by the crate before coming to rest. To find the distance, we'll use the equation \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration due to friction, and \(s\) is the distance traveled. \(0 = 8^2 + 2(-0.8)s\) \(s = \frac{8^2}{2(0.8)} = 40 \mathrm{m}\) Now that we have the distance, we can find the work done by friction: \(W = Fd = (40 \mathrm{N})(40 \mathrm{m}) = 1600 \mathrm{J}\).
06

Calculate the rate at which the frictional force takes energy away from the crate

The rate at which energy is taken away is the work done divided by the time taken to come to rest. \(\text{Energy Rate} = \frac{W}{t} = \frac{1600 \mathrm{J}}{10 \mathrm{s}} = 160 \mathrm{W}\). So the rate at which the frictional force takes energy away from the crate is \(160 \mathrm{W}\).

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Key Concepts

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

Kinematics
Kinematics is all about understanding the motion of objects without accounting for the forces that cause such motion. In this context, it involves analyzing how the crate's velocity changes over time as it moves across a rough surface. Let's break it down further:

- Initial velocity: The crate starts moving at a speed of 8 meters per second (m/s). This is the speed right before it experiences deceleration due to friction.
- Final velocity: The crate comes to a complete stop, meaning its final velocity is 0 m/s. This implies a change in velocity over time.

This motion can be described using the formula for acceleration: \[a = \frac{∆v}{t}\]where \(∆v\) is the change in velocity (which is final velocity minus initial velocity), and \(t\) is the time taken to stop.

In our problem, \(a\) is calculated as \(-0.8 \ \mathrm{m/s^2}\), indicating that friction is decelerating the crate over 10 seconds. Kinematics thus helps in calculating not only how quickly an object slows down but also the rate of this deceleration.
Work and Energy
Work and energy are closely linked concepts in physics. Here, we'll explore how the friction acting on the crate does work and thus removes energy from the moving crate. When we talk about work done by a force, we're interested in how much energy it transfers as it moves an object through a distance.

For our crate:
  • The frictional force (\(F\)) is calculated as 40 Newtons (N), acting in the opposite direction of the movement, hence decelerating the crate.
  • The distance the crate travels while decelerating is 40 meters (m). This is determined using kinematic equations based on initial velocity, final velocity, and acceleration.
The work done by friction, which is a measure of energy transferred, is given by:\[W = F \cdot d\]Substituting the given values, \(W = 40 \ \mathrm{N} \times 40 \ \mathrm{m} = 1600 \ \mathrm{Joules (J)}\).

This means friction takes 1600 Joules of energy away from the crate as it comes to a stop. The rate at which this energy is taken away can be calculated by dividing the work done (in Joules) by the time (in seconds), resulting in 160 Watts (W). This signifies how quickly the energy is drained during the deceleration process.
Newton's Laws
Newton's Laws of Motion, especially the second law, help explain why the crate slows down and stops. Newton's Second Law states that the acceleration of an object depends on the net force acting upon it and its mass, described by the equation \(F = ma\).

In this scenario, the crate has a mass of 50 kilograms (kg), and it undergoes an acceleration (or more precisely, deceleration) of \(-0.8 \ \mathrm{m/s^2}\) due to the frictional force exerted by the floor.

Calculating the frictional force involves replacing the variables in Newton's formula:
  • \(F = (50 \ \mathrm{kg}) \times (-0.8 \ \mathrm{m/s^2}) = -40 \ \mathrm{N}\)
The negative sign signifies that the force acts in the opposite direction to the crate's motion, hence slowing it down. This frictional force is crucial as it is responsible for the energy transfer, essentially transforming the crate's kinetic energy into heat, bringing the crate to a halt.

Newton's Laws provide a framework to understand how forces affect motion, which is key in comprehending various physical scenarios such as this one.

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Most popular questions from this chapter

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0 -kg sprinter running at \(10.0 \mathrm{m} / \mathrm{s}\) ? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

A boy pulls a 5-kg cart with a 20-N force at an angle of \(30^{\circ}\) above the horizontal for a length of time. Over this time frame, the cart moves a distance of \(12 \mathrm{m}\) on the horizontal floor. (a) Find the work done on the cart by the boy. (b) What will be the work done by the boy if he pulled with the same force horizontally instead of at an angle of \(30^{\circ}\) above the horizontal over the same distance?

(a) How long would it take a \(1.50 \times 10^{5}\) -kg airplane with engines that produce \(100 \mathrm{MW}\) of power to reach a speed of \(250 \mathrm{m} / \mathrm{s}\) and an altitude of \(12.0 \mathrm{km}\) if air resistance were negligible? (b) If it actually takes 900 s, what is the power? (c) Given this power, what is the average force of air resistance if the airplane takes 1200 s? (Hint: You must find the distance the plane travels in 1200 s assuming constant acceleration.)

Boxing gloves are padded to lessen the force of a blow. (a) Calculate the force exerted by a boxing glove on an opponent's face, if the glove and face compress 7.50 cm during a blow in which the 7.00 -kg arm and glove are brought to rest from an initial speed of \(10.0 \mathrm{m} / \mathrm{s}\). (b) Calculate the force exerted by an identical blow in the gory old days when no gloves were used, and the knuckles and face would compress only \(2.00 \mathrm{cm} .\) Assume the change in mass by removing the glove is negligible. (c) Discuss the magnitude of the force with glove on. Does it seem high enough to cause damage even though it is lower than the force with no glove?

Can the power expended by a force be negative?

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