A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40 and the coefficient of kinetic friction is 0.20.
If the monkey applies a horizontal force of 18.0 N, what is the box's acceleration?
(I tried to use newton's second law, F=ma and when I plugged the numbers in I got an acceleration of .45 m/s^2 but this is wrong)
If the monkey applies a horizontal force of 18.0 N, what is the box's acceleration?
(I tried to use newton's second law, F=ma and when I plugged the numbers in I got an acceleration of .45 m/s^2 but this is wrong)
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Taking g = 10m/s², the mass of the box = W/g = 40/10 = 4kg
Since the surface is horizontal, the normal force on the box, N = weight = 40N
The frictional force when moving = μk.N = 0.2 x 40 = 8N
The resultant horizontal force = 18 - 8 = 10N
Applying F =ma gives
10 = 4a
a = 2.50m/s² (give answer to 3 significant figures, same as the data supplied)
If you are supposed to use a more accurate value for g, it will give a slightly different answer.
Since the surface is horizontal, the normal force on the box, N = weight = 40N
The frictional force when moving = μk.N = 0.2 x 40 = 8N
The resultant horizontal force = 18 - 8 = 10N
Applying F =ma gives
10 = 4a
a = 2.50m/s² (give answer to 3 significant figures, same as the data supplied)
If you are supposed to use a more accurate value for g, it will give a slightly different answer.