Find the coefficient of friction

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Homework Statement



A box weighing 3.0 kN is placed in the bed of a pick-up truck. The pick-up accelerates uniformly from rest to 40 km/h in a distance of 40 m. In that time the box slides 1.0 m back toward the end of the bed. Find the coefficient of friction between the bed and the box.

Homework Equations



v^2 = v0^2 + 2a (x-x0)

F = ma

w = mg

Fk = micro k * N

The Attempt at a Solution



Given:
v0 = 0 km/h
v = 40 km/h
x-xo = Delta x = 40 m = 0.04 km
weight = 3 kN.

First, find a:

v^2 = v0^2 + 2 a (x-x0)
40^2 = 0 + 2 a * 0.04
=> a = 2*10^4 km/s^2.

My imagine that we have 3 forces in Free Body Diagrams: Friction Forces, Newton's law and weight of the box.

We have:

m*g = 3 kN => m = 3000/9.8 = 306.1 grams = 0.306 kg.

N = ma = 0.306 * (2*10^4) = 6120 kN

but the box slides 1.0 m back toward the end of the bed, so

(0.039km * 6120) / 0.04 = 5967 kN

Therefore:

coefficient of friction between the bed and the box is 5967 N / 6120 N = 0.975 = final answer => INCORRECT

Please help! Thanks!
 
Last edited:

Answers and Replies

  • #2
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My answer is NOT correct. Please help! Anyone?
 
  • #3
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Homework Statement



A box weighing 3.0 kN is placed in the bed of a pick-up truck. The pick-up accelerates uniformly from rest to 40 km/h in a distance of 40 m. In that time the box slides 1.0 m back toward the end of the bed. Find the coefficient of friction between the bed and the box.

Homework Equations



v^2 = v0^2 + 2a (x-x0)

F = ma

w = mg

Fk = micro k * N

The Attempt at a Solution



Given:
v0 = 0 km/h
v = 40 km/h
x-xo = Delta x = 40 m = 0.04 km
weight = 3 kN.

First, find a:

v^2 = v0^2 + 2 a (x-x0)
40^2 = 0 + 2 a * 0.04
=> a = 2*10^4 km/s^2.

My imagine that we have 3 forces in Free Body Diagrams: Friction Forces, Newton's law and weight of the box.

We have:

m*g = 3 kN => m = 3000/9.8 = 306.1 grams = 0.306 kg.

N = ma = 0.306 * (2*10^4) = 6120 kN

but the box slides 1.0 m back toward the end of the bed, so

(0.039km * 6120) / 0.04 = 5967 kN

Therefore:

coefficient of friction between the bed and the box is 5967 N / 6120 N = 0.975 = final answer => INCORRECT

Please help! Thanks!

Im no expert, i worked it out and this is what i got: You deduced the first part right, so
a = 1.5m/s2, then the weight of the box = 3000N, so mg = 3000N, so the force of friction = uk(3000N). If you use Newtons third law the foward force = the reaction force, so it is constant motion. So, the force of friction = Fnet, uk(3000N)=ma, a = the above acceleration and m is obtained from the weight equation Fg=mg, m = 306Kg, plug it into above equation and u shud get uk = 0.15.

Hope this helps:)
 
  • #4
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Im no expert, i worked it out and this is what i got: You deduced the first part right, so
a = 1.5m/s2, then the weight of the box = 3000N, so mg = 3000N, so the force of friction = uk(3000N). If you use Newtons third law the foward force = the reaction force, so it is constant motion. So, the force of friction = Fnet, uk(3000N)=ma, a = the above acceleration and m is obtained from the weight equation Fg=mg, m = 306Kg, plug it into above equation and u shud get uk = 0.15.

Hope this helps:)

How about "In that time the box slides 1.0 m back toward the end of the bed" ???
 
  • #5
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How about "In that time the box slides 1.0 m back toward the end of the bed" ???
Its neglected since the same acceleration is needed to move the box 1.0m.
 
  • #6
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Its neglected since the same acceleration is needed to move the box 1.0m.

You mean the final answer is 0.15 ???
 
  • #7
50
0
You mean the final answer is 0.15 ???

Yup, looks right:)
 

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