A 15.0-kg box is released from rest on a rough inclined plane tilted at an angle of 45° to the horizontal. The coefficient of kinetic friction between the box and the inclined plane is 0.200. What is the force of kinetic friction on the box?

Answer

Fifteen square root of two Newtons

Thirty square root of two Newtons.

Fifteen square root of three over two Newtons.

Fifteen square root of three Newtons.

Thirty square root of three Newtons.

Fn = mg*cos45 = 15 * 9.8 * cos45=104 N.

= Normal force = Force perpendicular to
the incline.

Fk = u*Fn = 0.2 * 104 = 20.8 N.

To find the force of kinetic friction on the box, we can use the formula:

Friction force = coefficient of kinetic friction * normal force

First, let's find the normal force. The normal force is the force exerted by the inclined plane perpendicular to its surface. In this case, since the box is on an inclined plane that is tilted at an angle of 45° to the horizontal, the normal force is equal to the weight of the box that acts perpendicular to the inclined plane.

The weight of the box can be calculated using the formula:

Weight = mass * acceleration due to gravity

Given that the mass of the box is 15.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight of the box:

Weight = 15.0 kg * 9.8 m/s^2 = 147 N

Since the inclined plane is tilted at 45° to the horizontal, the normal force is equal to the weight of the box, which is 147 N.

Now we can calculate the force of kinetic friction:

Friction force = 0.200 * 147 N = 29.4 N

So, the force of kinetic friction on the box is 29.4 Newtons.

Therefore, the correct answer is: Thirty square root of two Newtons.

To calculate the force of kinetic friction on the box, we need to use the formula:

Frictional force (Fk) = coefficient of kinetic friction (μk) * normal force (N)

First, let's calculate the normal force (N). The normal force is the perpendicular force exerted by a surface to support the weight of an object resting on it. In this case, the weight of the box is acting vertically downward, and the normal force acts perpendicular to the inclined plane.

The weight of the box (W) can be calculated using the formula:

Weight (W) = mass (m) * gravitational acceleration (g)

Given that the mass of the box is 15.0 kg and the gravitational acceleration is approximately 9.8 m/s^2, the weight of the box is:

W = 15.0 kg * 9.8 m/s^2 = 147 N

Since the inclined plane is tilted at an angle of 45° to the horizontal, we can decompose the weight into two components: one parallel to the inclined plane (W_parallel) and one perpendicular to the inclined plane (W_perpendicular).

W_parallel = W * sin(45°) = 147 N * (1/√2) = 103.9 N

W_perpendicular = W * cos(45°) = 147 N * (1/√2) = 103.9 N

Since normal force (N) is equal to the perpendicular component of weight (W_perpendicular), the normal force acting on the box is:

N = 103.9 N

Now that we have the normal force, we can calculate the force of kinetic friction:

Fk = μk * N = 0.200 * 103.9 N = 20.78 N (approximately)

Therefore, the force of kinetic friction on the box is approximately 20.78 Newtons.