A box with a mass of 10 kg is resting on a horizontal surface without friction.
ID: 1778018 • Letter: A
Question
A box with a mass of 10 kg is resting on a horizontal surface without friction. One force of 18 Newtons is applied straight downward. Another force of 5 Newtons is applied to the left. Lastly, a force is applied to it at an angle below the horizontal. (See sketch below.) The normal force on the box is 225 Newtons and after these forces have been applied when the box has slid 6 meters in the +x direction, the velocity of the box is 5 m/s in the +x direction. What angle (below the horizontal) was the third force applied in degrees?
Correct Asnwer: 76.65
Explanation / Answer
F1y = -18 N F1x = 0
F2x = -5 N F2y = 0
F3x = F3*costheta
F3y = -F3*sintheta
along perpendicular to the mass
n + W + F1y + F2y + F3y = 0
225 - 10*9.8 - 18 + 0 - F3*sintheta = 0
F3*sintheta = 109 N
along horizontal
acceleration a = (vf^2 - vi^2)/(2*x)
ax = (5^2-0)/(2*6) = 2.08 m/s^2
Fnet = F1x + F2x + F3x = m*a
-5 + F3x = 10*2.08
F3*costheta = 25.8 N
tantheta = 109/25.8
theta = tan^-1(109/25.8)
theta = 76.65 degrees
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