A 2 kg block is pushed with a force of 10 N against a horizontal surface. The block experiences a frictional force of 5 N opposing its motion. The coefficient of kinetic friction between the block and the surface is 0.4. The block initially at rest moves with an acceleration of 3 m/s². Determine:
a) The net force acting on the block. b) The force applied to overcome friction. c) The coefficient of static friction between the block and the surface.
a) The net force is equal to the mass of the block multiplied by its acceleration (according to Newton's second law). Thus, the net force is given by:
Net Force = mass × acceleration
Net Force = (2 kg) × (3 m/s²)
Net Force = 6 N
Answer: The net force acting on the block is 6 N.
b) The force applied to overcome friction is equal to the frictional force opposing the motion. Since the block experiences a frictional force of 5 N opposing its motion, the force applied to overcome friction is also 5 N.
Answer: The force applied to overcome friction is 5 N.
c) The coefficient of static friction (µ) can be determined using the equation:
µ = (Force of Friction) / (Normal Force)
Since the block is in motion, we can conclude that the force applied to overcome friction is equal to the force of friction. The normal force (N) is the force exerted by the surface on the block, which in this case is equal to the weight of the block acting vertically downwards.
Force of Friction = (Coefficient of kinetic friction) × (Normal Force)
5 N = (0.4) × N
N = 5 N / 0.4
N = 12.5 N
Therefore, the coefficient of static friction (µ) is given by:
µ = (5 N) / (12.5 N) = 0.4
Answer: The coefficient of static friction between the block and the surface is 0.4.