A mass of 8 kg is resting on a horizontal, frictionless surface. Force 1 is appl

Question

A mass of 8 kg is resting on a horizontal, frictionless surface. Force 1 is applied to it at 22 degrees above the horizontal, force 2 has a magnitude of 16 N and is applied vertically downward, force 3 has a magnitude of 8 N and is applied vertically upwards, and force 4 has a magnitude of 5 N and is applied in the -x direction to the object. If the amount of force 1 is such that the mass just begins to leave the surface, how long does it take the mass to move 22 meters in the +x direction in seconds?

Explanation / Answer

Just leave the surface means in vertical direction force should be balanced and normal due to surface should be zero. Net force vertically up = net force vertically down.

F1.Sin22 + F3 = F2 + Mg

Substituting values we get F1= 230.64 N

Net acceleration in horizontal direction = net force/M

= (230.64cos22 - 5)/8

= 26.105 m/se

Time taken to travel a distance of 22m =

t = (2×22/26.105)^(0.5)

= 1.298 sec

(I took g=9.8m/s2)

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