physics question

Question

A girl pushes a 2kg snow shovel at a uniform velocity across a sidewalk. If the

handle of the shovel is inclined at 55° to the horizontal and she pushes along the

handle with a force of 100 N, what is the force of friction? What is the coefficient of

kinetic friction?

Answer 1

F - mgsin55° - f = 0, where f is force of friction. So, f = F - mgsin55° = 100N - 2kg·9.81m/s²sin55° = 83.93 N

If μ is coefficient of kinetic friction, then f = μmgcos55°, μ = f/(mgcos55°) = 83.93/(2·9.81cos55°) = 7.46.

We get not real answer because such a data

Answer 2

Although the shovel is moving, it is not accelerating in any direction, as stated in the problem. This means that we can sum up our forces in both the x and y directions, and the sums of the forces will be equal to 0.

In the x-direction, we have the x-component of the force that the girls is applying to the shovel handle and the force of friction opposing it. In an equation, that looks like:

sumFx = Fgx-Ff = m*a, where the friction force (Ff) is an unknown, Fgx is the force of the girl in the x, and a=0.

sumFx = 100N*cos(55deg) - Ff = 0 --- here we can solve for the force of friction by moving it to the right side.

100N*cos(55deg) = Ff = 57.4N

As you can see, we did not need to use the sum of the forces in the y direction, as understanding this one simple equation was all we needed.

Hopefully this helps! Let me know if you have any follow-up questions!