a)
here
Ffriction = m * g * uk
Ffriction = 10 * 9.8 * 0.25 = 24.5 N
then the force that give motion to the block is = 30 - 24.5 = 5.5 N
then the acceleation of the block is
5.5 = m * a
a = 5.5 / 10 = 0.55 m/s^2
b)
here also we use the same formula
Ffriciton = 24.5 N
F = 90 - 24.5 = 65.5
then the acceleation is
a = 65.5 / 10 = 6.55 m /s^2
1. A sliding block of mass m-0.25 kg is subject to a force of magnitude 4 N that makes an angle of p 30° with the horizontal surface. If the coefficient of kinetic friction between block and surface is 0.5, what is the resulting acceleration of the block along the surface?
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