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anonymous
 5 years ago
2. A 2.0 kg block rests on a level surface. The coefficient of static friction is µs = 0.60, and the coefficient of kinetic friction is µk = 0.40. A horizontal force, F, is applied to the block. As F is increased, the block begins moving. Describe how the force of friction varies as F increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of F―before the block starts moving, at the point it starts moving, and after it is moving.
anonymous
 5 years ago
2. A 2.0 kg block rests on a level surface. The coefficient of static friction is µs = 0.60, and the coefficient of kinetic friction is µk = 0.40. A horizontal force, F, is applied to the block. As F is increased, the block begins moving. Describe how the force of friction varies as F increases from the moment the block is at rest to when it begins moving. Indicate how you could determine the force of friction at each value of F―before the block starts moving, at the point it starts moving, and after it is moving.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\mu = \frac{{F}f}{{F}n}\] Where u is the coefficient of friction, Ff is force of friction, and Fn is the normal force. The normal force is the force perpendicular (going straight up), so in this case, it is equal to the Force of Gravity on the block. 9.8 x 2 = 19.6 N, so this is the normal force. static friction is the friction when there is no motion, and kinetic friction is the friction when there is motion (the blocking moving) \[0.60 = \frac{{F}f}{19.6}\] for static friction \[0.40 = \frac{{F}f}{19.6}\] for kinetic friction So if we solve for both friction forces, we get that the static friction force is 11.8 N, and the kinetic friction force is 7.8 N. Before the block is moving, the force of friction is less than or equal to the static friction force (11.8 N), and when the block begins moving, the friction force is 7.8 N, the kinetic friction force.
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