# How much force is needed to give a 5.0kg box an acceleration of 0.2m/s^2 up a 30 degrees incline if the coefficient of friction is 0.30?

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by Jeffel

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by Jeffel

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The formula for this one is

acceleration = Force applied x (coefficient of force x normal force) - (mass x gravity x sine angle) / mass

**Note: Normal Force = mass x gravity x cosine angle**

Given:

Mass = 5 kg

Acceleration 0.2 m/s^2

Angle = 30 degrees

Coefficient of Friction = 0.30

Following the given what we are looking for here is the force applied so the formula would be like this.

**F = (gm sin(angle)) + am + nu**

where:

F = force applied

g = gravity

m = mass

u = coefficient of friction

n = normal force

F = ((9.8 m/s^2) x (5 kg) x sin 30) + ((0.2 m/s^2) x (5 kg)) + (((5 kg) x (9.8 m/s^2) x cos 30) x (0.30))

F = (24.5 N) + (1 N) + (42.435 N x 0.30)

F = (25.5 N) + (12.73 N)

**F = 38.23 N**

acceleration = Force applied x (coefficient of force x normal force) - (mass x gravity x sine angle) / mass

Given:

Mass = 5 kg

Acceleration 0.2 m/s^2

Angle = 30 degrees

Coefficient of Friction = 0.30

Following the given what we are looking for here is the force applied so the formula would be like this.

where:

F = force applied

g = gravity

m = mass

u = coefficient of friction

n = normal force

F = ((9.8 m/s^2) x (5 kg) x sin 30) + ((0.2 m/s^2) x (5 kg)) + (((5 kg) x (9.8 m/s^2) x cos 30) x (0.30))

F = (24.5 N) + (1 N) + (42.435 N x 0.30)

F = (25.5 N) + (12.73 N)