Question

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/s^2. 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?

Answer 1

a) To find the mass of the block, we can use the equation for force and acceleration:

Force = mass * acceleration

The force in this case is the applied force of 150 N, and the acceleration is 2.53 m/s^2. Plugging these values into the equation:

150 N = mass * 2.53 m/s^2

Rearranging the equation to solve for mass:

mass = 150 N / 2.53 m/s^2

mass ≈ 59.28 kg

Therefore, the mass of the block is approximately 59.28 kg.

b) When the block is sliding across the new surface at a constant speed, the applied force is balanced by the force of kinetic friction. The equation for this is:

Force of kinetic friction = coefficient of kinetic friction * normal force

Since the normal force is equal to the weight of the block (mg), and the applied force is 150 N, we can rewrite the equation as:

150 N = coefficient of kinetic friction * mg

We know from part a) that the mass of the block is 59.28 kg. The acceleration due to gravity is approximately 9.8 m/s^2. Plugging in these values:

150 N = coefficient of kinetic friction * 59.28 kg * 9.8 m/s^2

Simplifying the equation:

coefficient of kinetic friction = 150 N / (59.28 kg * 9.8 m/s^2)

coefficient of kinetic friction ≈ 0.267

Therefore, the coefficient of kinetic friction between the block and the new surface is approximately 0.267.