Work And Power Problems Using Inclined Planes and Forces

[beyondtheEND_]

Homework Statement

A)What is the minimum work needed to push a 1610 kg car 302 m up a 15 deg incline? Ignore friction.
B) Assume the effective coefficient of friction is 0.25.

(I have part A, the answer is 1.23x10^6 J)

Homework Equations

Force of Friction = [tex]\mu[/tex] x Normal Force
Work = Applied Force x cos([tex]\theta[/tex]) x distance

The Attempt at a Solution

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Homework Statement

A)A grocery cart with mass of 16.3 kg is pushed at constant speed along an aisle by a force of 10.4 N. The applied force acts at a(n) 22.3 deg angle to the horizontal. Find the work done by the applied force if the aisle is 14.9 m long.
B) Find the work done by the force of friction.
C) Find the work done by the normal force.
D) Find the work done by gravity.

(I have part A, the answer is 143.37 J)

Homework Equations

Force of Friction = [tex]\mu[/tex] x Normal Force
Work = Applied Force x cos([tex]\theta[/tex]) x distance
Work of Gravity = mass x gravity x height
Work of Normal Force = Normal Force x cos([tex]\theta[/tex])
Work of Force of Friction = Force of Friction x distance

The Attempt at a Solution

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Homework Statement

A)A 324-kg piano slides 5.16 m down a 30 deg incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of friction is 0.40. Calculate the force exerted by the man.
B) Find the work done by the man on the piano.
C) Find the work done by the force of friction.
D) Find the work done by gravity.
E) Find the net work done on the piano.

(I have part D, the answer is 8192.016 J)

Homework Equations

Force of Friction = [tex]\mu[/tex] x Normal Force
Work = Applied Force x cos([tex]\theta[/tex]) x distance
Work of Gravity = mass x gravity x height
Work of Normal Force = Normal Force x cos([tex]\theta[/tex])
Work of Force of Friction = Force of Friction x distance

The Attempt at a Solution

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I've been working on these problems for two days, nearly seven or eight hours in all, and yet I just can't seem to figure out what to do. I'm very sorry if you're not supposed to put more then one problem in each topic but they all use the same concepts so I thought this would be easiest. Any help someone could offer would be much appreciated! I understand the idea of what is going on and why things are happening, I simply get messed up with all the equations and the x-y tables. Please and thank you very much!

--beyondtheEND_​

 

[anathema]

Dear beyondtheEND_,

First of all, it is important to understand the concepts and equations involved in these problems. It seems like you have a good understanding of them, so that is a great start.

For part A of the first problem, the minimum work needed to push the car up the incline is equal to the change in potential energy of the car. This can be calculated using the equation PE = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the height the car is lifted. In this case, h can be calculated using the distance and the angle of the incline. So, the minimum work needed is equal to the change in potential energy, which is mgh. You can plug in the values for mass, height, and g to get the answer.

For part B of the first problem, you need to take into account the force of friction. The work done by the force of friction is equal to the force of friction multiplied by the distance traveled. The force of friction can be calculated using the equation F = μN, where μ is the coefficient of friction and N is the normal force. The normal force can be calculated using the equation N = mgcos(θ), where θ is the angle of the incline. Once you have the force of friction, you can multiply it by the distance traveled to get the work done by friction.

For part C of the first problem, the work done by the normal force is equal to the normal force multiplied by the distance traveled. The normal force can be calculated using the equation N = mgcos(θ), where θ is the angle of the incline. Once you have the normal force, you can multiply it by the distance traveled to get the work done by the normal force.

For part D of the first problem, the work done by gravity is equal to the weight of the car multiplied by the height it is lifted. This can be calculated using the equation W = mgh, where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s^2), and h is the height the car is lifted. In this case, h can be calculated using the distance and the angle of the incline. So, the work done by gravity is equal to the weight of the car multiplied by the height it is lifted.

For the second problem