Calculating Work for a 300kg Object on an Incline

In summary, we have a 300kg object moving down a 25 degree incline at a constant speed of 4.5 meters. It is kept from accelerating by a force pushing back on it, with an effective coefficient of friction of 0.39. The net work done on the object is 0, as it is not accelerating. The work done by the force pushing back on it is 917.1J, and the work done by gravity is 5597J. This problem can be solved by using the equations for gravity force, normal force, and friction force, and understanding the concept of net forces and work done.
  • #1
dzem68
3
0
Work Done?

If someone has the time and patience could you help me in setting this problem up?

300kg object moves 4.5 meters down a 25 degree incline at a constant speed. It is kept from accelerating by a force pushing back on it. The effective coefficient of friction is .39

I need to:

a) calculate the net work done on the object
b) calculate the work done by the force pushing back on it
c) calculate the work done by gravity on the object.


I do not know where to begin. could someone point me in the right direction.

thanks,

dz
 
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  • #2


Originally posted by dzem68
300kg object moves 4.5 meters down a 25 degree incline at a constant speed. It is kept from accelerating by a force pushing back on it. The effective coefficient of friction is .39

I need to:

a) calculate the net work done on the object
b) calculate the work done by the force pushing back on it
c) calculate the work done by gravity on the object.


gravity force:
Fg = 300 * 9.81
Fg = 2943N

normal force:
N = Fgcos(theta)
N = 2943cos(25)
N = 2667N

friction force:
f = uN
f = 0.39 * 2667
f = 1040N

Since the thing is not accelerating, net forces are 0. I drew a FBD of the box with the X axis along the 25 degree slant and the Y is the same direction as the normal force.
sum of x = 0 = 1040 + F - 2943sin(25)
F = 203.8N (the force preventing it from accelerating)


I don't exactly know what you mean by "net work". If I had to guess, I would say the net work is 0 since the mass is not accelerating.

work done by force:
W = Fd
W = 203.8 * 4.5
W = 917.1J

work done by gravity:
W = Fd
W = 2943sin(25) * 4.5
W = 5597J
 
Last edited:
  • #3
Thanks ShawnD! I read through your respons and I think I actually understand the concepts!
 

1. How do you calculate the work done on a 300kg object on an incline?

The work done on an object is equal to the force applied to the object multiplied by the distance it moves in the direction of the force. In this case, the force is the weight of the object, which is equal to its mass (300kg) multiplied by the acceleration due to gravity (9.8 m/s^2). The distance the object moves is the length of the incline. Therefore, the work done can be calculated as: Work = Force x Distance = (300kg x 9.8 m/s^2) x length of incline.

2. How does the angle of the incline affect the work done on a 300kg object?

The angle of the incline affects the work done because it determines the component of the object's weight that is acting in the direction of the incline. As the angle of the incline increases, the component of the weight acting in the direction of the incline decreases, resulting in less work being done on the object.

3. Can you use the same formula to calculate work for objects of different masses on an incline?

Yes, the same formula can be used to calculate work for objects of different masses on an incline. The only difference would be the weight of the object, which is equal to its mass multiplied by the acceleration due to gravity. As long as the force and distance are measured in the same units, the same formula can be used.

4. What is the unit of measurement for work?

The unit of measurement for work is the joule (J) in the International System of Units (SI). One joule is equal to the work done when a force of one newton is applied to an object and moves it one meter in the direction of the force.

5. Can you calculate the work done on an object on an incline if the angle of the incline is not given?

No, the angle of the incline is necessary to calculate the work done on an object. Without the angle, the component of the weight acting in the direction of the incline cannot be determined, and therefore the work cannot be calculated accurately.

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