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
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.
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.