A+block+of+1.2kg+is+placed+on+an+in cluded+plane+at+60°+to+the+horizont al.+Calculate.+a. The+force+that+will+make+the+block+ to+slide+down+the+plane. B.The+coefficient+of+friction+.C.+T he+normal+reaction+d.+The+friction+ force

a. 10.2 N b. 1.73 c. 58.8 N d. 10.2 N

Explanation:

a. The force that will make the block slide down the plane

This is the component of the block's weight along the plane F = mgsinθ where m = mass of block = 1.2 kg, g = 9.8 m/s² and θ = angle of incline = 60°.

So, F = mgsinθ = 1.2 kg × 9.8  m/s² × sin60°.= 10.18 N ≅ 10.2 N

b. The coefficient of friction

The coefficient of friction, μ = tanθ = tan60° = 1.73

c. The normal reaction.

This is equal to the vertical component of the block's weight. So, F = mgcosθ. Substituting the values for the variables from above, we have

F = mgcosθ = 1.2 kg × 9.8  m/s² × cos60°.= 58.8 N

d. The frictional force

Since the block does not slide, there is no net force on it. If f is the frictional force, then

F - f = ma. Since a = acceleration = 0,

F - f = 0

f = F = mgsinθ = 1.2 kg × 9.8  m/s² × sin60°.= 10.18 N ≅ 10.2 N

b

explanation: