Question

The 10 kg mass is released from rest at a height of 1m above the floor. If the co-efficient of kinetic friction between the 5 kg mass and the table is 0.28, what will be the speed of the 10 kg mass just before it hits the floor?

Please show the equation and then plug the numbers into that equation
V^2 = Vo^2 + 2g*h.
V^2 = 0 + 19.6*1 =
So the answer is 19.6 m/s right?
V^2 = 19.6, V = 4.43 m/s.

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