Question

A 2 kg block is sliding down a plane inclined at 30˚ to the horizontal. The...

A 2 kg block is sliding down a plane inclined at 30˚ to the horizontal. The coefficient of friction
          between the block and the plane is 0.25. How much work is done by gravity if the block slides 5 m?
          How much work is done by friction? If the block starts from rest, what is its speed after sliding the
          5 m?

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Answer #1

here

work done by the gravity

W = m * g * sin(theta) * d

W = 2 * 9.8 * sin30 * 5

W = 49 J

work done by friction force

W =u * m * g * cos30 * d

W = 0.25 * 2 * 9.8 * cos30 * -5

W = -21.22 J

then

Fnet = Fg - Ff

m * a = m * g * sin30 - u * m * g * cos30

a = g * (sin30 - u * cos30)

a = 9.8 (sin30 - 0.25 * cos30)

a = 2.78 m/s^2

then by using equation of law of motion

v^2 - u^2 = 2 * a * s

v = sqrt( 0 + 2 * 2.78 * 5)

v = 5.27 m/s

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