A box of bananas weighing 36.5 N restson a horizontal surface. The coefficient o
ID: 1662938 • Letter: A
Question
A box of bananas weighing 36.5 N restson a horizontal surface. The coefficient of static friction betweenthe box and the surface is 0.35, andthe coefficient of kinetic friction is 0.15.(a) If no horizontal force is applied to thebox and the box is at rest, how large is the friction force exertedon the box?
(b) What is the magnitude of the friction force if a monkey appliesa horizontal force of 5.5 N to the boxand the box is initially at rest?
(c) What minimum horizontal force must the monkey apply to startthe box in motion?
(d) What minimum horizontal force must the monkey apply to keep thebox moving at constant velocity once it has been started?
(e) If the monkey applies a horizontal force of 19.0 N, what is the magnitude of the frictionforce and what is the box's acceleration?
frictional force acceleration (a) If no horizontal force is applied to thebox and the box is at rest, how large is the friction force exertedon the box?
(b) What is the magnitude of the friction force if a monkey appliesa horizontal force of 5.5 N to the boxand the box is initially at rest?
(c) What minimum horizontal force must the monkey apply to startthe box in motion?
(d) What minimum horizontal force must the monkey apply to keep thebox moving at constant velocity once it has been started?
(e) If the monkey applies a horizontal force of 19.0 N, what is the magnitude of the frictionforce and what is the box's acceleration?
frictional force acceleration frictional force acceleration frictional force acceleration
Explanation / Answer
(a) If no horizontal force is applied to thebox and the box is at rest, how large is the friction force exertedon the box?if the box is at rest, friction force =0
(b) What is the magnitude of the friction force if a monkey appliesa horizontal force of 5.5 N to the boxand the box is initially at rest?
the max static friction force needed to move is .35*36.5 = 12.775N. since 5.5 N is less than that, the box is not going to move. sothe friction force will be 5.5N just enough to cancel the push.
(c) What minimum horizontal force must the monkey apply to startthe box in motion?
12.775 N
(d) What minimum horizontal force must the monkey apply to keep thebox moving at constant velocity once it has been started?
once it has started to move, you need to overcome kinetic frictiononly. force needed would be .15*36.5 = 5.475 N.
(e) If the monkey applies a horizontal force of 19.0 N, what is the magnitude of the frictionforce and what is the box's acceleration?
if the monkey applies 19 N and max. kinetic friction is 5.475 N,then
friction force = 5.475 N
a = (19-5.475)/(36.5/9.8) = 3.63 m/s^2
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