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a 5.00 kg stone is rubbed across the horizontal ceiling of a cave passageway (fig. 6-48). if the coefficient of kinetic friction is 0.65 and the force applied to the stone is angled at u ???? 70.0°, what must the magnitude of the force be for the stone to move at constant velocity?



Answer :

We take the +x direction to be horizontal to the right and the +y direction to be up.

Newton’s second law

The equations for the x, y components of the force as per Newton’s second law are:

                  Fx​=Fcosθ−f=ma

                  Fy​=Fsinθ−FN​−mg=0

Now, f=μk​FN​ and the second equation from above gives FN​=Fsinθ−mg, which yields f=μk​(Fsinθ−mg).

This expression is replaced for f in the first equation to obtain

                  Fcosθ−μk​(Fsinθ−mg)=ma.

For a = 0 , the force is F= cosθ− μk​sinθ− μk​mg​

With μk​=0.65, m = 5.0kg, and θ = 70∘, we obtain F = 118 N.

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