Question

A block of mass 7.50 kg is observed at rest on a 25.0° slope. From this observation, what can be concluded about the friction coefficient between the block and the surface of the slope?

Answer 1

If the block of mass 7.50 kg is observed at rest on a 25.0° slope, it means that the force of friction acting on the block is equal and opposite to the component of the gravitational force that is parallel to the slope. This can be expressed as:

f_friction = m*g*sin(25°)

where f_friction is the force of friction, m is the mass of the block, g is the acceleration due to gravity (9.81 m/s^2), and sin(25°) is the sine of the angle of the slope.

The coefficient of friction between the block and the surface of the slope can be calculated using the equation:

μ = f_friction / f_normal

where μ is the coefficient of friction and f_normal is the normal force acting on the block, which is equal to the component of the gravitational force that is perpendicular to the slope. This can be expressed as:

f_normal = m*g*cos(25°)

where cos(25°) is the cosine of the angle of the slope.

Substituting the values of f_friction and f_normal into the equation for the coefficient of friction, we get:

μ = (m*g*sin(25°)) / (m*g*cos(25°))

Simplifying, we get:

μ = tan(25°)

Using a calculator, we find that:

μ ≈ 0.4663

Therefore, the coefficient of friction between the block and the surface of the slope is approximately 0.4663.