T and Pushing Box - o F N1S = m*a o 300N*10m/s 2 o 3000 N...

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25). A fish (8 kg) is being yanked vertically upward out of the water and the fishing line breaks. If the line is rated to a maximum tension of 240 N (~ 53 lb test), then what was the minimum acceleration of the fish (in m/s 2 )? T + W1E = m*a 240kg + -80 N W1E = 8kg * a a= 20 m/s 2 26). A crate (40 kg) is hung in an elevator by a cord rated to withstand a tension of 700 N. The elevator starts at rest and begins moving upwards until the cord holding the crate breaks. What was the acceleration of the elevator when the cord broke? Assume g = 10 m/s 2 . Multiply the weight of the crate to find force in Newtons. o 40kg*10 m/s 2 = 400 N Subtract the force you just calculated from the withstanded tension. This gives you the net force. o 700N – 400N = 300 N Then divide net force by the weight of the crate. o 300 N / 40 kg o 7.5 m/s 2 27). A box (300 kg) is on the floor. The coefficient of static friction = 0.8 between the box and the surface of the shelf. Assume g = 10 m/s 2 . What is the normal force on the box due to the surface of the shelf?
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Unformatted text preview: o F N1S = m*a o 300N*10m/s 2 o 3000 N What is the maximum amount oF Force that can be exerted on the box due to static Friction? o F NIS * μ= F max o 3000 N* 0.8 o 2400 N The box is pushed with a force of 2350.0 N. What is the amount of force exerted on the box due to static friction? o The static friction force is equal to the opposing applied force o F μ =|-2350 N| o 2350 N The static friction will continue to increase until the box is pushed with a force of 2400.0 N. At this point the box will begin to move and therefore the motion of the box will now be opposed by kinetic friction. If the coeFcient of kinetic friction = 0.3, what is the magnitude of the acceleration ? o μ – μ f 0.8-0.3 = 0.5 o Use 0.5 and multply by F N1S 3000 N * 0.5 1500 N o Divide The max amounT of force exerTed based o± of new μ To geT The magniTude of The acceleraton. 1500 N / 300kg 5 m/s 2...
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