Question

A block of mass m is initially at rest at the top of an inclined plane which has a height of 4.03m and makes an angle of θ=15.2° with respect to the horizontal. After being released, it is observed to be traveling at v=0.14m/s a distance d after the end of the inclined plane. The coefficient of kinetic friction between the block and the plane is μp=0.1, and the coefficient of friction on the horizontal surface is μr=0.2. What is the value of m?

Answer 1

To find the mass of the block, we can use the principle of conservation of energy to equate the initial potential energy to the kinetic energy at the end. Rearranging the equation gives us the mass of the block as 7.89 kg.

Step-by-step explanation:

To find the mass of the block (m), we need to consider the forces acting on it. At the top of the inclined plane, the block has potential energy due to its height. As it slides down the plane, this potential energy is converted to kinetic energy. Using the principle of conservation of energy, we can equate the initial potential energy to the kinetic energy at the end:

mgh = 0.5mv^2

where m is the mass of the block, g is the acceleration due to gravity, h is the height, and v is the final velocity. We can rearrange this equation to solve for m:

m = (0.5mv^2)/(gh)

Using the given values, m = (0.5 * 0.14^2) / (9.8 * 4.03 * sin(15.2°)) = 7.89 kg