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a sled weighing 300 N is moved at constant speed over a horizontal floor by a force of 50 N applied parallel to the floor
determine the coefficient of friction between the sled and the floor

Respuesta :

MonsieurPengu
If the sled is moving at a constant speed, that mean there needs to be an equal but opposite force countering the 50 N force. This opposite force is friction, and is equal to the weight times a factor:

F = n*W

Let n be the coefficient of friction.

Now you simply fill in the formula and you get:

50 N = n*300N or n = 50N / 300N = 1/6

So, the final answer is 1/6 or 0.1667
snehashish65

The coefficient of friction between the sled and the floor is 0.167.

Given data:

The weight of Sled is, W = 300 N.

The magnitude of applied force is, F = 50 N.

For the constant speed motion, the applied force should overcome frictional force acting between the sled and the floor. Then,

Frictional force = Applied force

[tex]f = F\\\mu \times N = F[/tex]

Here, N is the normal reaction force, which is acting perpendicular on the sled, and numerically its value is equal to the weight. Then,

N = W

Solving as,

[tex]\mu \times N = F\\\mu \times W = F\\\mu \times 300 = 50\\\mu =0.167[/tex]

Thus, we can conclude that the coefficient of friction between the sled and the floor is 0.167.

Learn more about the coefficient of friction here:

https://brainly.com/question/12213966