Question

A 200kg block slides down a surface inclined at 30 degrees

The coefficient of kinetic friction between the block and the surface is 0.1
What is the magnitude of the friction force?
(acceleration due to gravity: 9.8ms−2)

Answer 1

First, we need to determine the force of gravity acting on the block. We can do this using the formula:

force of gravity = mass x acceleration due to gravity

force of gravity = 200kg x 9.8ms^-2

force of gravity = 1960N

Next, we need to determine the component of the force of gravity that is acting parallel to the surface. We can do this using the formula:

force parallel to surface = force of gravity x sin(angle of incline)

force parallel to surface = 1960N x sin(30 degrees)

force parallel to surface = 980N

Finally, we can determine the friction force using the formula:

friction force = coefficient of friction x force perpendicular to surface

The force perpendicular to the surface is equal to the force of gravity acting perpendicular to the surface, which can be calculated using the formula:

force perpendicular to surface = force of gravity x cos(angle of incline)

force perpendicular to surface = 1960N x cos(30 degrees)

force perpendicular to surface = 1695.7N

Now we can calculate the friction force:

friction force = 0.1 x 1695.7N

friction force = 169.57N

Therefore, the magnitude of the friction force acting on the block is 169.57N.