A box is being pushed up the ramp by you at a 15 degree angle. The coefficient of kinetic friction between the box and ramp is .4. The box has a mass of 12 kg and you are pushing up the ramp with a force of 95N. What is the kinetic frictional force? What is the sum of forces (net force) acting parallel to the plan?

for the first question I got -46N and on the second question do I add gravity parallel to the plane with frictional force and add that to force push?

To determine the kinetic frictional force, we can use the formula:

F_friction = coefficient of kinetic friction * normal force

The normal force is the perpendicular force exerted by the ramp on the box, which is equal to the weight of the box since it is on an 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 12 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can find the weight:

Weight = 12 kg * 9.8 m/s^2 = 117.6 N

The normal force is then equal to the weight, so:

Normal force = 117.6 N

Now, we can plug in the given coefficient of kinetic friction, which is 0.4, and solve for the frictional force:

F_friction = (0.4)(117.6 N) = 47 N.

Therefore, the kinetic frictional force is 47 N.

To find the net force acting parallel to the plane, we need to consider all the forces acting in that direction.

First, let's calculate the force component parallel to the plane due to the force you are pushing with:

Force parallel to plane (from pushing) = force * sin(angle)

Given that the force you are pushing with is 95 N and the angle of the ramp is 15 degrees:

Force parallel to plane (from pushing) = 95 N * sin(15°) = 24.58 N

Next, let's find the force component parallel to the plane due to gravity:

Force parallel to plane (gravity) = weight * sin(angle)

Since we have already calculated the weight to be 117.6 N:

Force parallel to plane (gravity) = 117.6 N * sin(15°) = 30.40 N

Finally, the net force acting parallel to the plane is the sum of these two forces:

Net force (parallel to the plane) = Force parallel to plane (from pushing) + Force parallel to plane (gravity)
= 24.58 N + 30.40 N
= 54.98 N

Therefore, the sum of forces (net force) acting parallel to the plane is approximately 54.98 N.

To find the kinetic frictional force, you can use the equation:

Frictional force = coefficient of kinetic friction * normal force

The normal force can be calculated by considering the forces acting perpendicular to the ramp. Since the box is on a ramp inclined at a 15-degree angle, the normal force can be found using:

Normal force = mass * gravitational acceleration * cos(angle)

Here, the gravitational acceleration is 9.8 m/s^2.

Substituting the given values, we have:

Normal force = 12 kg * 9.8 m/s^2 * cos(15 degrees) ≈ 114.96 N

Now, calculating the kinetic frictional force:

Frictional force = 0.4 * 114.96 N ≈ 45.98 N

Therefore, the kinetic frictional force acting on the box is approximately 45.98 N.

Now, moving on to the second question, we need to find the net force acting parallel to the ramp. This involves considering the individual forces acting in that direction.

The force you are applying is 95 N, acting up the ramp. The force of kinetic friction is acting in the opposite direction, so it can also be considered negative (since friction opposes motion). Hence, the kinetic frictional force is -45.98 N.

The gravitational force acting parallel to the ramp can be calculated using:

Gravitational force parallel to the ramp = mass * gravitational acceleration * sin(angle)

Substituting the given values:

Gravitational force parallel to the ramp = 12 kg * 9.8 m/s^2 * sin(15 degrees) ≈ 30.23 N

Now, to find the net force acting parallel to the ramp, we need to add up all the individual forces in that direction:

Net force parallel to the ramp = Force you are applying + Gravitational force parallel to the ramp + Kinetic frictional force

Plugging in the values:

Net force parallel to the ramp = 95 N + 30.23 N - 45.98 N ≈ 79.25 N

Therefore, the sum of forces (net force) acting parallel to the ramp is approximately 79.25 N.