How to Calculate Work on an Inclined Surface?

AI Thread Summary
To calculate the work done on an inclined surface, the formula W = f(cos θ)d is used, where θ is the angle of the force and d is the distance. In the first scenario, a man pulls a 10 kg box with a force of 87 Newtons at an angle of 24 degrees, resulting in 6040 joules of work. For the second part, the box is pulled up an incline at a 10.6-degree angle at constant speed, indicating that the work done equals the change in gravitational potential energy. The vertical displacement is calculated as 1.83 meters using trigonometric principles. The work done against gravity is then determined by the formula W = mg*h, where m is mass, g is acceleration due to gravity, and h is the height.
Iamaskier721
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Homework Statement


A man pulls a 10 kg box across a smooth floor with a force of 87 Newtons at an angle of 24 degrees and for a distance of 76 meters. How much work, to the nearest joule, does he do?

If the floor in the previous question is angled upward at 10.6 degrees and the man pulls the box up the floor at constant speed what is the work he does to the nearest joule?


Homework Equations



W=f(cos theta)d

The Attempt at a Solution


The first part i achieved and got correct, which is 6040 j. I have no idea how to do the second part. Please help!
 
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Since the speed is not changing, the work is equal to the change in gravitational potential energy. Actual distance doesn't matter, only the change in height.
 
So it is being displaced 1.83 m veritcally. I Figured this out by the law of sines and a simple right triangle. What is the next step to solving this?
 
Iamaskier721 said:
So it is being displaced 1.83 m veritcally. I Figured this out by the law of sines and a simple right triangle. What is the next step to solving this?

The work going into the change in gravitational potential is the weight times the height. In this case it is mg*h.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
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Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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