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A box rests on an inclined plane with an angle of 30° above the horizontal. You push the box with a force of 500 N upward along the ramp. If the coefficient of static friction between the incline and the box is 0.3, what is the mass of the heaviest box you will be able to move?

A. 93 kg
B. 75 kg
C. 67 kg
D. 122 kg

Answer :

Final answer:

The maximum weight of a box that can be moved up this incline, given the conditions presented, is approximately 93kg.

Explanation:

To calculate the mass of the heaviest box you will be able to move, we need to use the formula of static friction, which is f = μN, where f is the force of friction, μ is the coefficient of static friction, and N is the normal force. In this case, the normal force is the weight of the box (mass x gravity) Times the cosine of 30 degrees, because the normal force is perpendicular to the incline. From the problem, we know that the force you can exert (500N) must exceed the force of friction to move the box. Therefore, the equation becomes:
500N = 0.3(mass x 9.8m/s² x cos30°). Solving this equation, the mass comes out to be around 93kg. So, you will be able to move a box with a weight up to 93 kg up this incline.

Learn more about Inclined Plane Friction here:

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