Ch 09: Work and Kinetic Energy

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 09: Work and Kinetic Energy

Problem 64

Chapter 9, Problem 64

A farmer uses a tractor to pull a 150 kg bale of hay up a 15° incline to the barn at a steady 5.0 km/h. The coefficient of kinetic friction between the bale and the ramp is 0.45. What is the tractor's power output?

Verified step by step guidance

Step 1: Identify the forces acting on the bale. These include the gravitational force (mg), the normal force, the frictional force, and the force exerted by the tractor. The gravitational force can be broken into two components: one parallel to the incline (m * g * sin(θ)) and one perpendicular to the incline (m * g * cos(θ)).

Step 2: Calculate the normal force. The normal force is equal to the perpendicular component of the gravitational force, which is given by: F_normal = m * g * cos(θ). Use the mass of the bale (150 kg), the gravitational acceleration (g = 9.8 m/s²), and the incline angle (15°).

Step 3: Determine the frictional force. The frictional force is given by: F_friction = μ_k * F_normal, where μ_k is the coefficient of kinetic friction (0.45). Substitute the value of the normal force calculated in Step 2 to find F_friction.

Step 4: Calculate the total force required to pull the bale up the incline at a constant velocity. Since the bale is moving at a steady speed, the net force is zero. Therefore, the tractor's force must balance the sum of the parallel component of gravity (m * g * sin(θ)) and the frictional force (F_friction).

Step 5: Compute the power output of the tractor. Power is given by: P = F_total * v, where F_total is the total force calculated in Step 4 and v is the velocity of the bale (5.0 km/h, converted to m/s). Substitute the values to find the power output.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Power

Power is defined as the rate at which work is done or energy is transferred over time. In physics, it is calculated as the work done divided by the time taken. The unit of power is the watt (W), where 1 watt equals 1 joule per second. Understanding power is essential for determining how much energy the tractor uses to pull the bale up the incline.

Work and Energy

Work is the product of the force applied to an object and the distance over which that force is applied, in the direction of the force. In this scenario, the work done by the tractor includes overcoming gravitational forces and friction as it pulls the bale up the incline. The concept of energy conservation is also relevant, as the work done translates into kinetic and potential energy changes.

Friction

Friction is the resistive force that opposes the motion of an object in contact with a surface. The coefficient of kinetic friction quantifies this resistance and is crucial for calculating the force needed to move the bale up the incline. In this problem, the frictional force must be considered alongside gravitational forces to determine the total force exerted by the tractor.

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Related Practice

Textbook Question

Astronomers using a 2.0-m-diameter telescope observe a distant supernova - an exploding star. The telescope's detector records 9.1 x 10^-11 J of light energy during the first 10 s. It's known that this type of supernova has a visible-light power output of 5.0 x 10³⁷ W for the first 10 s of the explosion. How distant is the supernova? Give your answer in light years, where one light year is the distance light travels in one year. The speed of light is 3.0 x 10⁸ m/s.

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Textbook Question

Write a realistic problem for which this is the correct equation(s).

$T - (1500 kg) (9.8 {m/s}^{2}) = (1500 kg) (1.0 {m/s}^{2})$

$P equals T of open paren 2.0 m slash s close paren$

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Textbook Question

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. How long does it take the Porsche to reach the maximum power output?

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Textbook Question

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. If the Porsche accelerates at a_max, what is its speed when it reaches maximum power output?

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Textbook Question

Draw a pictorial representation.

$T-(1500\,$\text{kg}$)(9.8\,$\text{m/s}$^2)=(1500\,$\text{kg}$)$\left$(1.0\,$\text{m/s}$^2$\right$)$

$P=T(2.0\,$\text{m/s}$)$

Textbook Question

A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the engine and the drive train, and 70% reaches the wheels. The total mass of the car and driver is 1480 kg, and two-thirds of the weight is over the drive wheels. What is the maximum acceleration of the Porsche on a concrete surface where μ_s = 1.00? Hint: What force pushes the car forward?

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