How much work is done by friction in this scenario?

[Hughey85]

Hi everyone. I have another problem that is just stumping me. I posted this earlier and still can't understand it. Here it is...

A 1-kg ball starting at h = 6.1 meters slides down a smooth surface where it encounters a rough surface and is brought to rest at B, a distance 18.3 meters away. To the nearest joule what is the work done by friction?

Imagine the ball starting on top of a waterslide, sliding down with no friction, and then coming down onto the straight path that slows you down. I can find the Potential Energy of the ball, but don't know where to go from there! Pls help!

 

[Leong]

Law of conservation of energy :
Total energy in the beginning=Total energy in the end
$$mgh= W_f$$
The potential energy of the ball has been transformed to heat etc due to the work done by friction.

 

[wais]

In this scenario, the work done by friction can be calculated by using the formula W = Fd, where W is work, F is the force of friction, and d is the distance over which the force is applied. Since the ball is sliding down a smooth surface at the beginning, there is no friction and therefore no work is done. However, once the ball encounters the rough surface, friction comes into play and slows down the ball, bringing it to rest at point B.

To calculate the work done by friction, we need to first find the force of friction. This can be done by using the formula F = μN, where μ is the coefficient of friction and N is the normal force (equal to the weight of the ball, which is 1 kg in this case). The coefficient of friction depends on the roughness of the surface and can range from 0 (no friction) to 1 (maximum friction).

Next, we need to find the distance over which the force of friction is applied. This is equal to the distance between the starting point (h = 6.1 meters) and the point where the ball comes to rest (B = 18.3 meters). Therefore, d = 18.3 - 6.1 = 12.2 meters.

Now, we can plug in the values into the formula W = Fd. The force of friction is μN = μ(1 kg)(9.8 m/s^2) = 9.8μ N. Plugging this into the formula, we get W = (9.8μ N)(12.2 m) = 119.56 μ J. Since we are asked to round to the nearest joule, the work done by friction in this scenario is approximately 120 J.

I hope this helps! Remember, the key to solving problems like this is to understand the concepts and formulas involved, and then use them to apply to the specific scenario given. Good luck!