Question

A 2.00-kg block on a 30° inclined plane slides down at a constant speed of 3.00...

A 2.00-kg block on a 30° inclined plane slides down at a constant speed of 3.00 m/s. What is the power of the frictional force in moving the block?

Homework Answers

Answer #1

Given that block is moving with constant speed, So that means acceleration of block is zero, and net force on block along the incline is zero.

Now Using force balance along the incline:

F_net = W*sin - Ff

Since F_net = 0, So

W*sin - Ff = 0

Ff = Friction force = W*sin

Given that: W = Weight of block = m*g = 2.00*9.81 N

= incline angle = 30 deg

So,

Ff = 2.00*9.81*sin 30 deg = 9.81 N

Friction force between block and ramp = 9.81 N

Now power of frictional force will be given by:

Power = Friction force*Velocity

Power = 9.81*3.00

Power = 29.43 W

Let me know if you've any query.

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