a block of mass m  5.00 kg ispulled along a horizontal frictionless floor by a cordthat exerts a force of magnitude F = 12.0 N at anangle u = 25.0. (a) What is the magnitude of theblock’s acceleration? (b) The force magnitude F isslowly increased. What is its value just before theblock is lifted (completely) off the floor? (c) What isthe magnitude of the block’s acceleration just before it is lifted(completely) off the floor?

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A: The net force  is 5.60 So, 5.60 = ACos32+Bcos32 As, A=B So, 5.6=2A×0.848 This implies A=5.6/1.696…

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Q: o that Fair = c v, c

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A: By fbd of block  MgSin20° + F - ukMgCos20° = Ma  F = M(a + ukgCos20° - gSin20°)  F = 1900 ( 5 +…

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Q: In the figure, a block of mass m = along a horizontal frictionless floor by a cord that exerts a…

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A: Given :  vo = 50.0 km/h = 50.0(1000/3600) = 13.9 m/s v = 0 m/s (because finally the car stops)

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Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”