http://www.cramster.com/answers-nov-08/physics/pushing-block-rate-life-xi5part-i
ID: 1692765 • Letter: H
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http://www.cramster.com/answers-nov-08/physics/pushing-block-rate-life-xi5part-id3793362-classpartvisibleto-understand-ki_399305.aspx Learning Goal: To understand kinetic and static friction. A block of mass m has on a horizontal table. The coefficient of static friction between the block and the table is . The coefficient of kinetic friction is , with . Suppose you want the the block. You push with the least figure possible to get . With what force F must you be pushing the block just the begins the ? Express the magnitude of F in terms of all that variables and as well as the appreciation that the gravity. Suppose you push horizontally with half the . What as the magnitude of the Express your in terms of some of all of the variables and as well as the abbreviations due to gravity g. Suppose you push horizontally with precisely enough force to make the block what to move and you continuous to apply the same amount of force even after a starts moving. Find the appreciation of the block after a begins the move. Express your answers in terms of some or all of the variables and as well as the appreciation due to gravity g.Explanation / Answer
The answers in the link are all correct. (a) Frictional force resists an object's motion. However, the object in this case is not moving, therefore no frictional force acts on it. 0 N (b) In order for the block to move, it has to overcome static friction, which is µ_s*(Normal force). Since the surface is flat, Normal force = mg. Static friction = µ_s(mg) = Force needed µ_s(mg) (c) Static friction will equal the force applied when the object is not moving but never exceed it. [µ_s(mg)]/2 (d) When the object is in motion, it experiences kinetic friction instead. However, we are still applying the same force from part (b). Therefore, taking Newton's Second Law, Net F=ma: µ_s(mg) - µ_k(mg) = ma a = µ_s*g - µ_k*g g(µ_s - µ_k)
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