A 400 N block is dragged along a horizontal surface by an applied force. THe coefficient of kinetic friction is Uk= 0.4 and the block moves at constant velocity. The magnitude of vector F is:
I know the answer is 150, I need to know how to solve that.
I know the answer is 150, I need to know how to solve that.
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Because the velocity is constant, you also know that the acceleration in that direction is zero. Therefore, using the Sum(Forces)=Mass*acceleration:
F - Uk*F(normal) = 0
F = Uk*F(normal)
It's horizontal, so your Fn=400N. Uk=.4
400*(.4)=F=150 N
And you're done. If this block was on an inclined plane, you'd have to take into account that the normal force is actually mg*cos(theta), and therefore your Uk*Fn would be Uk*(mg*cos(theta)).
F - Uk*F(normal) = 0
F = Uk*F(normal)
It's horizontal, so your Fn=400N. Uk=.4
400*(.4)=F=150 N
And you're done. If this block was on an inclined plane, you'd have to take into account that the normal force is actually mg*cos(theta), and therefore your Uk*Fn would be Uk*(mg*cos(theta)).