A1250-kg slippery hippo slides down a mud-covered hill inclined at an angle of 18º to the
horizontal. if the coefficient of sliding friction between the hippo and the mud is 0.0900, what
force of friction impedes the hippo’s motion down the hill?

The solution to the problem is as follows:

The force of friction can be calculated using the equation F = uN, where F is the magnitude of the force, u is the coefficient of friction, and N is the normal force. For an inclined plane at 18 degrees above the horizontal, the normal force is given by N = mg cos theta. Putting all the equations together,

F = umg cos theta
F = (0.09)(1250)(9.80) cos(18)
F = 1.05 kN

I hope my answer has come to your help. Thank you for posting your question here in . We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

25% increaseplus i think decrease i only know increase: ((