Question

A wooden block slides directly down an inclined plane, at a constant velocity of 6.0 m/s. a) How large is the coefficient of kinetic friction if the plane makes an angle of 25 degrees with the horizontal? b) If the angle of incline is changed to 10 degrees, how far will the block slide before coming to a stop?

Answer 1

To find the coefficient of kinetic friction, use the formula frictional force = coefficient of kinetic friction * normal force. When the angle of incline is changed, use force of kinetic friction = coefficient of kinetic friction * normal force to calculate the distance the block will slide.

Step-by-step explanation:

To find the coefficient of kinetic friction, we can use the formula:

frictional force = coefficient of kinetic friction * normal force

Since the block is sliding down the inclined plane at a constant velocity, the frictional force is equal to the force component parallel to the incline. This force can be calculated using:

force parallel to incline = mass * acceleration parallel to incline

Using the given values, we can calculate the coefficient of kinetic friction as:

coefficient of kinetic friction = frictional force / (mass * acceleration parallel to incline)

For part b), when the angle of incline is changed to 10 degrees, the block will slide until the force of kinetic friction exactly cancels out the component of gravity pulling the block down the incline. This force is given by:

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

We can calculate the distance the block will slide using:

distance = (velocity^2) / (2 * acceleration)

Answer 2

• a) The coefficient of kinetic friction for the wooden block sliding down an inclined plane at an angle of 25 degrees is approximately 0.36.
• b) The block will slide approximately 3.1 meters before coming to a stop at an angle of 10 degrees.

Step-by-step explanation:

a) In order to find the coefficient of kinetic friction, we need to use the formula:

f`k = µk mg cos θ

where f`k is the frictional force, µk is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of the incline.

Plugging in the values given in the question:

θ = 25°, m = 2 kg, and f`k = 4.86 N

we can solve for µk:

4.86 N = µk * 2 kg * 10 m/s² * cos(25°)

µk ≈ 0.36

Therefore, the coefficient of kinetic friction is approximately 0.36.

b) To find how far the block will slide before coming to a stop at an angle of 10 degrees, we can use the work-energy principle:

W = ∆KE + ∆PE + ∆U

where W is the work done on the block, KE is the kinetic energy, PE is the potential energy, and U is the work done by the frictional force. Since the block comes to a stop, its final kinetic energy is zero. The change in potential energy is given by ∆PE = mgh, where h is the vertical height. The work done by the frictional force is given by ∆U = -f`k * d * cos(θ), where d is the distance the block slides and θ is the angle of the incline.

Plugging in the values given in the question:

θ = 10°, m = 2 kg, and f`k ≈ 0.36 (from part a)

we can solve for d:

0 = 0 + m * g * h + (-0.36) * d * cos(10°)

d ≈ 3.1 m

Therefore, the block will slide approximately 3.1 meters before coming to a stop at an angle of 10 degrees.