Determine the coefficient of friction

Feb 2020
8
2
Sweden
Hi I need help with the following question:


A small box with the mass of 5 kg is sent with an inital speed of 5.6 m/s up an sloping plane with an incline angle of 20 °. After 2.0 m the box stops. Determine the coefficient of friction between the box and the plane.


First attempt:

Suppse the box coverts its kinetic energy \(\displaystyle W_{k} \) into work of friction \(\displaystyle W_{f} \) where \(\displaystyle W_{f} = F_{k}S \) and \(\displaystyle F_{k} \) denotes the force of kinetic friction and \(\displaystyle S \) the displacement. Then it follows that

\(\displaystyle W_{k} = W_{f} \iff (1/2)(mv^2)=F_{k}S \iff F_{k} = \frac{mv^2}{2S}\approx 39,2 N\) . We know that \(\displaystyle \mu_{k}= \frac {F_{k}}{F_{N}} \) and by trigonomety \(\displaystyle F_{N}=-mgcos(20^{\circ}) \) therefore

\(\displaystyle \mu_{k} = \frac{\frac{mv^2}{2S}}{-mgcos(20^{\circ})}=\frac{mv^2}{-2Smgcos(20^{\circ})}=\frac{v^2}{-2Sgcos(20^{\circ})}=\frac{(5.6)^{2}}{4(9.82)cos(20^{\circ})}\approx 0.8496\)

Second attempt:

The resultant force \(\displaystyle F_{∥ }\) is parallel to the contact surface. We know that the sum of forces acting on the box parallel to the contact surface is equal to \(\displaystyle F_{∥ } =Fg_{∥} + F_{k} \) where \(\displaystyle Fg_{∥} \) denotes the component force of gravity with is parellel to the contact surface and\(\displaystyle F_{k} \) the force of kinetic friction. From trigonomety \(\displaystyle Fg_{∥}=Sin(20^{\circ}) Fg=Sin(20^{\circ}) mg\) and the normal force \(\displaystyle F_{N}=cos (20^{\circ})mg\). By definition \(\displaystyle F_{k} =\mu_{k}(F_{N}) =\mu_{k}cos (20^{\circ})mg\)


Since \(\displaystyle F_{∥ }=Fg_{∥} + F_{k} =Sin(20^{\circ})mg +\mu_{k}Cos(20^{\circ})mg = mg(Sin(20^{\circ})+\mu_{k}Cos(20^{\circ}))=a(m)\). Then the acceleration of the box is equal to

\(\displaystyle a = g(Sin(20^{\circ})+\mu_{k}Cos(20^{\circ}))\). From kinematics we can calculate \(\displaystyle a \) by using the formula \(\displaystyle v_{f}^{2}=v_{i}^{2}+2as\) given that \(\displaystyle v_{i}=5.6 \space m/s \), \(\displaystyle v_{f}=0 \space m/s \) and \(\displaystyle s =2 \space m\). Then \(\displaystyle a = -\frac{v_{i}^{2}}{2s}= -\frac{5.6^{2}}{4}=-7.84\). Now we can find \(\displaystyle \mu_{k} \) given that \(\displaystyle g=-9.82 \space m/s^{2} \) thus \(\displaystyle \mu_{k} =\frac{a-gSin(20^{\circ})}{gCos(20^{\circ})} \approx 0.4856 \)
 
Last edited:
Apr 2015
1,156
303
Somerset, England
Do you have something against diagrams?

That's a lot of maths to wade through.

The trick with friction-on-a-slope questions is to be able to determine which way the frictional force acts.


Friction always opposes the (actual or potential) direction of motion.

So depending upon what is holding the object to the plane, friction can act either up or down the plane.
 
Feb 2020
8
2
Sweden
I don't mind using diagrams, however I don't know how to draw diagrams on a forum. I'm aware that one normally define \(\displaystyle F_{∥ }= -Fg_{∥} - F_{k} \) assuming \(\displaystyle g\) is positive. And I know that \(\displaystyle Fg_{∥}\) and \(\displaystyle F_{k}\) are the forces which oppose the directed motion of the object.
 
Last edited:
Apr 2015
1,156
303
Somerset, England
First attempt

If you are going to use energy methods you need also to account for the increase in gravitational potential energy of the block.

so you have initial kinetic energy = energy of work against friction + increase in PE of block.
 
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Feb 2020
8
2
Sweden
First attempt

If you are going to use energy methods you need also to account for the increase in gravitational potential energy of the block.

so you have initial kinetic energy = energy of work against friction + increase in PE of block.
Thanks now I know what I did wrong!