Solving Physics Problems with the Box and Banana

AI Thread Summary
The discussion focuses on solving physics problems related to a box of bananas weighing 40.0N on a horizontal surface, addressing static and kinetic friction. The friction force is zero when no horizontal force is applied, and a 6.0N force from a monkey results in a friction force equal to the applied force. To initiate motion, the monkey must apply a minimum force of 16N, while maintaining constant velocity requires 8N. When the monkey exerts an 18.0N force, the net force acting on the box is 10N, leading to an acceleration of 2.45m/s² after correcting for mass. The calculations and friction forces are confirmed to be accurate for the given scenarios.
EvanQ
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Homework Statement



A box of bananas weighing 40.0N 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.

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?

b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0N to the box and the box is initially at rest?

c) What minimum horizontal force must the monkey apply to start the box in motion?

d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started?

e) If the monkey applies a horizontal force of 18.0N, what is the magnitude of the friction force ?

f) If the monkey applies a horizontal force of 18.0N, what is the box's acceleration?


Homework Equations



F=ma

The Attempt at a Solution



a) if there is no force applied to the box, the friction force exerted is 0N? or would i times the coefficients of both static and kinetic energy with the weight of the box?

b) just the 6N applied? or the (static friction coefficient x weight) - 6N?

c) 40N x coefficient of static = 16N

d) 40N x coeff of kinetic = 8N

e) remains 8N

f) having a major mind blank over this one.

so yeh, just needing some pointers for parts a, b and f.
thanks in advance.
 
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just got part b correct, only need help with part a and f thanks.
 
For a you're right, there is 0 frictional force.

For f... what is the net force acting in the horizontal direction while the box is moving? Use this and F=ma to get the acceleration.
 
18N to the right, minus 8N to the left (kinetic friction x weight) = 10N to the right.

F=ma
10=40a
a=0.25m/s?

or am i using the wrong value for m?
 
EvanQ said:
18N to the right, minus 8N to the left (kinetic friction x weight) = 10N to the right.

F=ma
10=40a
a=0.25m/s?

or am i using the wrong value for m?

The m is wrong. 40N is the weight not the mass. Get the mass from W = mg.
 
once again on my final submission :p, so could you please just verify this is correct for me>

F=ma
10=4.08a
a=2.45m/s

thanks
 
EvanQ said:
once again on my final submission :p, so could you please just verify this is correct for me>

F=ma
10=4.08a
a=2.45m/s

thanks

Yup, only the units are wrong. :wink: 2.45m/s^2
 
sweet thanks :)
 
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