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A force of 32 n is required to start a 6 kgbox moving across a horizontal concrete floor. what is the coefficient of static friction between the box and the floor?

User Cfr
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Final answer:

The coefficient of static friction between the box and the horizontal concrete floor is approximately 0.544, calculated using the maximum static frictional force and the weight of the box.

Step-by-step explanation:

To calculate the coefficient of static friction between the box and the horizontal concrete floor, we use the formula fs(max) = μsN, where fs(max) is the maximum static frictional force that needs to be overcome to start moving the object, μs is the coefficient of static friction, and N is the normal force (which is equal to the weight of the object if on a horizontal surface). In this case, the maximum static frictional force required to start moving the box is given as 32 N, and the normal force is the weight of the 6 kg box. Since the box is on a horizontal surface, its weight W is equal to mg, where m is the mass and g is the acceleration due to gravity (9.8 m/s²). The weight of the box is therefore (6 kg)(9.8 m/s²) = 58.8 N. Now, we can solve for the coefficient of static friction using the formula: μs = fs(max)/N. Plugging in the values we have μs = 32 N / 58.8 N ≈ 0.544. The coefficient of static friction between the box and the horizontal concrete floor is approximately 0.544.

User Dktaylor
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