Question

an 11 kg block slides down a surface with friction that is inclined at an angle of 38 degrees. how much work does the friction force do after the block has gone .75 m down the incline? .275 friction

Answer #1

Work-done by Friction Force will be:

W = F.d = F*d*cos

F = Friction Force on block = *N

N = m*g*cos

d = displacement of block along the incline = 0.75 m

= Coefficient of kinetic friction = 0.275

m = mass of block = 11 kg

= Incline angle = 38 deg

= 180 deg = Angle between friction force and displacement (Since friction force is always in opposite direction of motion)

So,

W = (*m*g*cos )*d*cos

Using given values:

W = (0.275*11*9.81*cos 38 deg)*0.75*cos 180 deg

**W = -17.54 J**

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