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A still block slides over a still surface,if the mass of the pot is 2.00kg what is the force of friction opposing the motion?coefficient of kinetic friction is taken as 0.42.Gravity is 9.8

User Icon for Henry Henry answered
12 years ago

Wb = mg = 2.0kg * 9.8N/kg = 19.6N. =

Weight of block.

Fb = 19.6N @ 0deg.
Fv = 19.6cos(0) = 19.6N. = Force perpendicular to the surface.

Fk = u*Fv = 0.42*19.6 = 8.23N. = Force of kinetic friction.

User Icon for Explain Bot Explain Bot answered
11 months ago

To determine the force of friction opposing the motion, we can use the following formula:

Frictional Force (Ff) = Coefficient of Kinetic Friction (μ) * Normal Force (N)

First, let's calculate the normal force. Since the block is on a still surface and there is no vertical acceleration, the normal force is equal to the weight of the block.

Weight (W) = mass (m) * gravity (g)

Given:
Mass (m) = 2.00 kg
Gravity (g) = 9.8 m/s^2

Weight (W) = 2.00 kg * 9.8 m/s^2 = 19.6 N

Now, we can substitute the values into the formula:

Frictional Force (Ff) = 0.42 * 19.6 N

Calculating the frictional force:

Ff = 0.42 * 19.6 N = 8.232 N

Therefore, the force of friction opposing the motion is approximately 8.232 N.