Question #269376

Jasmine is cycling at 12 m s−1 when her bag falls off the back of her bicycle.The bag slides a distance of 9 m before coming to rest. Calculate the coefficient of friction between the bag and the road.


1
Expert's answer
2021-11-23T13:00:04-0500

The work-energy theorem states that the work done on an object by the net force is equal to the change in its kinetic energy:


mv222mv122=Ffrs\dfrac{mv_2^2}{2}-\dfrac{mv_1^2}{2}=-F_{fr}\cdot sFfr=μmgF_{fr}=\mu m g

mv222mv122=μmgs\dfrac{mv_2^2}{2}-\dfrac{mv_1^2}{2}=-\mu mg\cdot s

μ=v12v222gs\mu=\dfrac{v_1^2-v_2^2}{2gs}

Given v1=12 m/s,v2=0 m/s,s=9m,g=9.81 m/s2.v_1=12\ m/s, v_2=0\ m/s, s=9m, g=9.81\ m/s^2.


μ=(12 m/s)2(0 m/s)22(9.81 m/s2)(9 m)=0.8155\mu=\dfrac{(12\ m/s)^2-(0\ m/s)^2}{2(9.81\ m/s^2)(9\ m)}=0.8155

The coefficient of friction between the bag and the road is 0.8155.



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