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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?

Sagot :

Otakuprinceidn
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