A box with mass 11.0 kg moves on a ramp that is inclined at an angle of 55.0 abo
ID: 1411831 • Letter: A
Question
A box with mass 11.0 kg moves on a ramp that is inclined at an angle of 55.0 above the horizontal. The coefficient of kinetic friction between the box and the ramp surface is k = 0.300.
(a) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 150.0 N that is parallel to the ramp surface and directed down the ramp, moving the box down the ramp.
(b) Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 150.0 N that is parallel to the ramp surface and directed up the ramp, moving the box up the ramp.
Explanation / Answer
Newton's II law:
F = ma = F(net)
also Ffr = kinetic frictional force = N [N is the normal reaction of the surface on the box = mgcos55°]
A.
along the incline equation is:
F + mgsin55° - Ffr = ma
=>150.0 + 11gsin55° - 11gcos55° = ma
=> 238.30 - 18.568 = 11a
=> a = 19.97 m/s² (down the ramp)
B.
equation is: F - mgsin55° - Ffr = ma
=> 150 - 11gsin55° - 11gcos55° = ma
a = 5.60 m/s² (up the ramp)
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