Answer:
μk = 0.3
Explanation:
- According the work-energy theorem, the net work done on an object, is equal to the change in the kinetic energy of the object.
- In this case, assuming that the object was not accelerating when moving at 8.0 m/s, the only force doing work on it, is the kinetic friction force, acting through the 11 m distance before causing the object to come to rest.
- The kinetic friction force, is given by the following expression:
[tex]f_{frk} = \mu_{k} * F_{n} (1)[/tex]
- where μk = coefficient of kinetic friction, Fn = normal force.
- Assuming that the surface is horizontal, since the normal force is always perpendicular to the surface, and the object is not accelerating in the vertical direction, this means that the normal force must be equal and opposite to the force that gravity exerts on the object, as follows:
[tex]F_{n} = m*g (2)[/tex]
- The friction force and the horizontal displacement produced by it have opposite directions, so the angle between the force and the displacement is 180º.
- So, we can express the work done by the kinetic friction force, as follows:
[tex]W_{ffr} = -F_{fr} * d = - \mu_{k}* m*g*d (3)[/tex]
- We have already said that (3) must be equal to the change in the kinetic energy, ΔK, as follows:
[tex]\Delta K = K_{f} - K_{o} (4)[/tex]
- Since the object comes to rest, Kf = 0.
- Replacing K₀ = 1/2*m*v₀² in (4), we have:
[tex]\Delta K = 0 - \frac{1}{2} *m * v_{o} ^{2} = - \frac{1}{2} *m * v_{o} ^{2} (5)[/tex]
- From (3) and (5), since they are equal each other, we get:
[tex]W_{ffr} = \mu_{k}* g*d = \frac{1}{2} * v_{o} ^{2} (6)[/tex]
- Replacing by the givens, and solving for μk, we get:
[tex]\mu_{k} = \frac{1}{2} * v_{o} ^{2} *\frac{1}{g*d} =\frac{(8.0m/s)^{2}}{2*9.8m/s2*11m} = 0.3 (7)[/tex]
