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A 0.5 kg block is being slid up a chalkboard with an applied force of 6.0 N upward and 2.0 N inward towards the board. If the coefficient of kinetic friction is 0.4.

a) Calculate the force of friction.

b) Calculate the acceleration of the block.

F = mju R [ F = u R]

1. F/u = 6/0.4 = 15N ?

2. F = ma ... 6/0.5 = a = 12m/s^2

not sure... some1 back me up on this!

a) Calculate the force of friction.

b) Calculate the acceleration of the block.

F = mju R [ F = u R]

1. F/u = 6/0.4 = 15N ?

2. F = ma ... 6/0.5 = a = 12m/s^2

not sure... some1 back me up on this!

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