/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 A loaded ore car has a mass of \... [FREE SOLUTION] | 91Ó°ÊÓ

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A loaded ore car has a mass of \(950 \mathrm{~kg}\) and rolls on rails with negligible friction. It starts from restand is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at \(30.0^{\circ}\) above the horizontal. The car accelerates uniformly to a speed of \(2.20 \mathrm{~m} / \mathrm{s}\) in \(12.0 \mathrm{~s}\) and then continues at constant speed. (a) What power must the winch motor provide when the car is moving at constant speed? (b) What maximum power must the motor provide? (c) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length \(1250 \mathrm{~m} ?\)

Short Answer

Expert verified
The power provided by the winch motor when the car is moving at a constant speed is about \(16.4 \, \mathrm{kW}\). The maximum power provided by the motor is about \(16.8 \, \mathrm{kW}\). The total energy transferred out of the motor by work by the time the car moves off the end of the track is about \(9321.5 \, \mathrm{kJ}\).

Step by step solution

01

Calculate the force exerted by the winch

Firstly, we need to find the force exerted by the winch for the car to accelerate uniformly. We use the kinematic equation \(v = u + at\), where \(v\) is the final speed, \(u\) is the initial speed (0 in this case because the car starts from rest), and \(t\) is the time. Solving for acceleration (a) gives \(a = (v-u)/t = 2.2/12.0 = 0.183 \, \mathrm{m/s^2}\). The force exerted by the cable should be equal to the weight component along the incline plus the force required to accelerate the car on the inclined plane. The weight component along the incline is \(m \cdot g \cdot \sin(\theta)\) and the force required to provide the acceleration is \(m \cdot a\). So, \(F = m \cdot g \cdot \sin(\theta) + m \cdot a\). Substituting the known values \(m = 950 \, \mathrm{kg}\), \(g = 9.8 \, \mathrm{m/s^2}\), \(\theta = 30.0^{\circ}\), we get \(F = 950 \cdot 9.8 \cdot \sin(30) + 950 \cdot 0.183 = 7455.35 \, \mathrm{N}\).
02

Calculate the power when the car is moving at a constant speed

The power is the work done per unit time. At constant speed, the work done by the force is the force times the velocity along the direction of the force. So, the power \(P = F \cdot v = 7455.35 \cdot 2.2 = 16381.77 \, \mathrm{W}\) or \(16.4 \, \mathrm{kW}\). Note that the work done is resistive and does not contribute to the kinetic energy of the car.
03

Calculate the maximum power provided by the winch

The maximum power occurs when the car is accelerating and the work done by the winch not only overcomes the resistive force but also contributes to the kinetic energy of the car. Therefore, the maximum power is the sum of the power computed in step 2 and the power due to acceleration. The power due to acceleration is \(P_{acc} = m \cdot a \cdot v = 950 \cdot 0.183 \cdot 2.2 = 383.67 \, \mathrm{W}\). Thus, the maximum power \(P_{max} = P + P_{acc} = 16381.77 + 383.67 = 16765.44 \, \mathrm{W}\) or \(16.8 \, \mathrm{kW}\).
04

Calculate the total energy transferred by work

The work done by the winch motor as the car moves along the track is the work done to overcome the resistive force and accelerate the car. The work done to move the car a distance \(d = 1250 \, \mathrm{m}\) against the resistive force is \(W_{res} = F \cdot d = 7455.35 \cdot 1250 = 9319187.5 \, \mathrm{J}\). The work done to accelerate the car is \(W_{acc} = \frac{1}{2} m \cdot v^2 = 0.5 \cdot 950 \cdot (2.2)^2 = 2321 \, \mathrm{J}\). The total energy transferred is \(W_{total} = W_{res} + W_{acc} = 9319187.5 + 2321 = 9321508.5 \, \mathrm{J}\) or \(9321.5 \, \mathrm{kJ}\).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that explores the motion of objects without considering the forces causing the motion. It focuses mainly on parameters such as displacement, velocity, and acceleration. In the given problem, we encounter kinematics when calculating the acceleration of the ore car. The car starts from rest and accelerates to a speed of 2.20 m/s over 12 seconds. To determine its acceleration, we use the kinematic equation: \[ v = u + at\]Here, \( v \) is the final velocity, \( u \) is the initial velocity (which is 0 since the car starts from rest), \( a \) is the acceleration, and \( t \) is the time. By solving for \( a \), we find that \( a = 0.183 \mathrm{~m/s^2} \). Kinematics helps us understand how an object's position or velocity changes over time.
Work and Energy
Work and energy are fundamental concepts in physics, often described as the ability to do work. The concept of work involves a force causing displacement. In this context, the work is done by the winch motor pulling the ore car along the inclined plane. When the car accelerates, a portion of the energy goes into increasing its kinetic energy and overcoming resistive forces. Work done is calculated by the formula: \[W = F \times d\]Here, \( W \) is the work done, \( F \) is the force applied, and \( d \) is the displacement. For our ore car, two types of work are relevant: the work to overcome resistive forces \( (W_{res}) \) and the work to provide acceleration \( (W_{acc}) \).- Resistive work: \( W_{res} = 7455.35 \times 1250 \) Joules- Acceleration work: \( W_{acc} = \frac{1}{2} \times 950 \times (2.2)^2 \) JoulesThe total energy transferred is the sum of both: \( 9321508.5 \) Joules.
Inclined Planes
An inclined plane is a flat surface tilted at an angle compared to the horizontal, often used to simplify the lifting of heavy objects. An inclined plane allows for the reduction of the force needed to lift something by extending the distance over which the force is applied. In our problem, the inclined plane is angled at \( 30^{\circ} \) above the horizontal. This angle affects both the gravitational force component acting along the incline and the work done by the winch motor. The force along the incline is given by:\[F_{\text{incline}} = m \times g \times \sin(\theta)\]Where \( m \) is the mass (950 kg), \( g \) is the gravitational acceleration (9.8 \mathrm{~m/s^2}), and \( \theta \) is the incline angle. The inclined plane reduces the effective force required to pull the car upwards compared to a direct lift, making the work more manageable for the winch motor.
Power Calculations
Power is the rate at which work is done or energy is transferred. It is measured in watts, with one watt being equivalent to one joule per second. In this ore car scenario, power is essential for determining the capacity of the winch motor needed to pull the car effectively.When the car moves at a constant speed: \[P = F \times v\]Where \( P \) is the power, \( F \) is the force, and \( v \) is the velocity. The calculated power when the car travels at constant speed is about 16.4 kW.The maximum power, when the car is accelerating, adds the power needed to overcome inertia: \[P_{\text{max}} = P + m \times a \times v\]This requires adding the power needed for acceleration, resulting in a maximum power need of 16.8 kW. Understanding power helps in determining how quickly work is done and the potential limits of machinery, like the winch motor in our problem.

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Most popular questions from this chapter

A block of mass \(12.0 \mathrm{~kg}\) slides from rest down a frictionless \(35,0^{\circ}\) incline and is stopped by a strong spring with \(k=3.00 \times 10^{4} \mathrm{~N} / \mathrm{m} .\) The block slides \(3.00 \mathrm{~m}\) from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring been compressed?

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?

In a circus performance, a monkey is strapped to a sled and both are given an initial speed of \(4.0 \mathrm{~m} / \mathrm{s}\) up a \(20^{\circ}\) inclined track. The combined mass of monkey and sled is \(20 \mathrm{~kg}\), and the coefficient of kinetic friction between sled and incline is \(0.20\). How far up the incline do the monkey and sled move?

A light spring with spring constant \(1.20 \times 10^{3} \mathrm{~N} / \mathrm{m}\) hangs from an elevated support. From its lower end hangs a second light spring, which has spring constant \(1.80 \times 10^{3} \mathrm{~N} / \mathrm{m}\). A \(1.50-\mathrm{kg}\) object hangs at rest from the lower end of the second spring. (a) Find the total extension distance of the pair of springs. (b) Find the effective spring constant of the pair of springs as a system. We describe these springs as being in series. Hint: Consider the forces on each spring separately.

A \(0.250\) -kg block along a horizontal track has a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) immediately before colliding with a light spring of force constant \(4.60 \mathrm{~N} / \mathrm{m}\) located at the end of the track. (a) What is the spring's maximum compression if the track is frictionless? (b) If the track is not frictionless, would the spring's maximum compression be greater than, less than, or equal to the value obtained in part (a)?

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