A box of bananas weighing 38.5 N rests on a horizontal surface. The coefficient

Question

A box of bananas weighing 38.5 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 0.23. (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 6.4 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/s^2

Explanation / Answer

(a) In the absence of External Force, frictional force is ZERO

(b) Ans: 6.4 N

Maximum value of Static Friction is 0.35 x 38.5 = 13.475 N

Now, since External Force applied is only 6.4 N, Hence friction present will also be 6.4 N

(c) Ans 13.475 N

Minimum Horizontal Force to apply = Maximum value of Static Friction = 13.475 N

(d) Ans: 8.855 N

After start of motion , monkey should apply Force equal to Kinetic Friction, to continue it in constant velocity

Kinetic Friction = 0.23 x 38.5 = 8.855 N

(e) Applied Force = 19 N

Frictional Force = Kinetic Friction = 8.855 N

Net Force = 19 - 8.855 = 10.145 N

acceleration = Net Force /mass = 10.145 x 9.8/38.5 = 2.58 m/s2

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