A box of bananas weighing 38.0 N rests on a horizontal surface. The coefficient
ID: 1536989 • Letter: A
Question
A box of bananas weighing 38.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.35, and the coefficient of kinetic friction is 0.18.
(a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? ___ N
(b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.5 N to the box and the box is initially at rest? ___ N
(c) What minimum horizontal force must the monkey apply to start the box in motion? ___ N
(d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? ___ N
(e) If the monkey applies a horizontal force of 19.0 N, what is the magnitude of the friction force and what is the box's acceleration?
frictional force ___ N acceleration ___ m/s2
Explanation / Answer
mass = 38/9.8 = 3.877 kg
a) Force of friction = 0.35 ( 38) = 13.3 N
b) frictional force remaining = 13.3 - 6.5= 6.8 N
c) minimum force that must be applied to start the box= 13.3 N
d)frictional force under motion = 0.18 ( 38) = 6.84 N ...this must force must be appled to keep the body under constant velocity
e)magnitude of frictional force = 6.84 N
net force = applied force - frictional force
net forc e= 19 - 6.84= 12.16 N
a= 12.16/ 3.877 = 3.136 m/s^2 apprx
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