Coefficient of kinetic friction and pulling force

[eureka360]

hi everybody.. can i ask how to get a coefficient of kinetic friction if a 20 kilogram sled is being pulled across a horizontal surface at a constant velocity. The pulling force has a magnitude of 80 Newton and is directed at an angle of 30 degrees above the horizontal. please answer my question.. i badly need you help..

 

[gunblaze]

I assume you know the formula for finding kinetic friction. If the sled is moving at constant velocity, what will be the relationship between the horizontal component of the pulling force and the kinetic friction? How do you find the normal force given the weight of the sled and the upward force that the person is helping ease the contact between the sled and the floor.

 

[kyle8921]

Hello,

To calculate the coefficient of kinetic friction, we need to use the equation Ff = μkN, where Ff is the force of friction, μk is the coefficient of kinetic friction, and N is the normal force (equal to the weight of the sled, in this case).

In this scenario, the pulling force of 80 Newton is equal to the force of friction, since the sled is moving at a constant velocity. We can rearrange the equation to solve for μk: μk = Ff/N = 80N/(20kg x 9.8m/s^2) = 0.4082.

It is important to note that the coefficient of kinetic friction is a unitless quantity and can vary depending on the surface and materials involved. It is also important to consider that the angle of the pulling force may affect the normal force and therefore the coefficient of kinetic friction.

I hope this helps. If you have any further questions, please let me know.