Question

A 2kg block of a material slides down a 30 degree and 2m inclined plane with a consistent acceleration. The block starts from rest at the top and gained its velocity of 5m/s. What is the acceleration of the block and coefficient of the friction?

Answer 1

The acceleration of the block is 5 m/s^2 and the coefficient of friction is 0.2472.

Step-by-step explanation:

To find the acceleration of the block, we can use the formula a = gsin(theta), where g is the acceleration due to gravity (10 m/s^2) and theta is the angle of the incline (30 degrees). So, a = 10 * sin(30) = 5 m/s^2. The force of the ramp on the block can be found using the formula F = mgcos(theta), where m is the mass of the block (2 kg) and g is the acceleration due to gravity. So, F = 2 * 10 * cos(30) = 17.32 N. To find the force applied upward along and parallel to the ramp that would allow the block to move with constant velocity, we can use the formula F = mkN, where N is the normal force and mk is the coefficient of kinetic friction. Since the block is moving with constant velocity, the frictional force is equal to the force applied upward. So, 4.86 N = mk * 19.62 N (mg), mk = 0.2472.