A block is pushed across a horizontal surface with a coefficient of kinetic friction of 0.15 by applying a
150 N horizontal force.
a) The block accelerates at the rate of 2.53 m/s2
. Find the mass of the block.
b) The block slides across a new surface while experiencing the same applied force as before. The
block now moves with a constant speed. What is the coefficient of kinetic friction between the
block and the new surface?
a) To find the mass of the block, we can use Newton's second law of motion, which states that the force applied to an object is equal to the mass of the object multiplied by its acceleration:
F = ma
In this case, the force applied to the block is 150 N and the acceleration is 2.53 m/s^2. Rearranging the equation, we can solve for the mass:
m = F/a = 150 N / 2.53 m/s^2 = 59.28 kg
Therefore, the mass of the block is 59.28 kg.
b) When the block is sliding across the new surface with a constant speed, it is experiencing a force of kinetic friction that is equal to the force applied to the block (150 N) in the opposite direction. The force of kinetic friction is given by:
fk = μk * N
where μk is the coefficient of kinetic friction and N is the normal force. Since the block is moving with a constant speed, the force of kinetic friction is equal in magnitude and opposite in direction to the applied force:
fk = 150 N
Now, we can rearrange the equation to solve for the coefficient of kinetic friction:
μk = fk / N
The normal force is equal in magnitude to the force of gravity acting on the block:
N = mg
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Therefore, μk = 150 N / (mg) = 150 N / (59.28 kg * 9.8 m/s^2) = 0.260
Therefore, the coefficient of kinetic friction between the block and the new surface is approximately 0.260.