A 1.40-kg object slides to the right on a surface having a coefficient of kineti
ID: 2238127 • Letter: A
Question
A 1.40-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.80 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e). Find the distance of compression d. This would be the correct answer if there were no force of friction. m Find the speed v at the unstretched position when the object is moving to the left (Figure d). How much energy is stored in the spring just before the object makes contact with the spring? How much energy is stored in the spring after the object leaves the spring? Do you need to consider the spring potential energy in working this part? m/s Find the distance D where the object comes to rest. This part is easiest to work by starting with your answer to part (b) (if it is correct!), but it can also be worked using only your answer to part (a). To understand how, think about the initial state of the system (object traveling with a speed 2.80, spring uncompressed) and the final state of the system (object at rest, spring uncompressed). What happens to the kinetic energy of the object? mExplanation / Answer
a) conservation of energy
Ei + W friction = Ef
1/2 mv^2 - u m g d = 1/2 k d^2
0.5*1.4*2.8^2 - 0.25*1.4*9.81*d = 0.5*50*d^2
d=0.405 m
b) Ei + W friction = Ef
1/2 mv^2 - um g 2d = 1.2 mv^2
0.5*1.4*2.8^2 - 0.25*1.4*9.81*2*0.405 = 0.5*1.4*v^2
v=1.966 m/s
c) Ei + W friction = Ef
1/2 mv^2 - um g (2d+D) = 1.2 mv^2
0.5*1.4*2.8^2 - 0.25*1.4*9.81*(2*0.405+D) =0
D=0.788 m
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