A box (200 kg) is on the floor. The coefficient of static friction = 0.8 between

Question

A box (200 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 box What is the normal force on the box due to the surface of the shelf? Submit Answer Tries 0/2 What is the maximum amount of force magnitude of the force that can be exerted on the box due to static friction? Submit Answer Tries 0/2 The box is pushed with a force of 1550.0 N. What is the amount of force magnitude of the force exerted on the box due to static friction? Submit Answer Tries o/2 The static friction will continue to increase until the box is pushed with a force of 1600.0 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 coefficient of kinetic friction = 06, what is the magnitude of the acceleration ? m/s2 Submit Answer Tries o/2

Explanation / Answer

N = m g = 200 x 10 = 2000 N .....Ans

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f_max = us N = 0.8 x 2000 = 1600 N .....Ans

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F_applies is less than f_max.

so f = F_applied = 1550 N ......Ans

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fk = 0.6 x 2000 = 1200 N  

Fnet = m a

1600 - 1200 = 200 a

a = 2 m/s^2 ......Ans

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