A 1.20-kg object slides to the right on a surface having a cofficient of kinetic

Question

A 1.20-kg object slides to the right on a surface having a cofficient of kinetic friction 0.250 (Fiqure a). The object has a speed o v-3.50 my's when t makes contact with a ight sering spring (Figure d) and continaes to move is that constant of s0.0 N/m. The object comes to rest after the spring has been compressed a distance d (igure c). The object is then forced toward the left by the the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (igure e). (a) Find the distance of compression d 4866 (b) Find the speed v at the unstretched position when the object is moving to the left (Figured How much energy is stored in the spring just before the object makes contact with the spring? How much energy is stored in the soring after the object leaves the sprig? Do you need to spring potential energy in working this part? m/s comes to rest (c) Find the distance D where the object to part (b) (if it is correcti), but it can also be worked using only your ansaer to part (a) To understand how, think about the lital state of the est to work by starting with your answer essed and the final state of the sstem fohtert at

Explanation / Answer

a]

Using conservation of energy

(1/2)mv2 - ukmgd = (1/2)kd2

=> (1/2)(1.2)(3.5)2 = 0.25(1.2 x 9.8)d + 25d2

Solve this quadratic equation to get:

d = 0.4866 m

b] When the spring decompresses, the elastic potential energy = 0 J and so using conservation of energy, the final kinetic energy when the object is moving to the left will be:

K.Ef = (1/2)mv2 = (1/2)mu2 - ukmg(2d)

[the factor of 2 comes in d because the object travels the distance d two times: once when the spring compresses and another when the spring decompresses].

=> (1/2)(1.2)v2 = (1/2)(1.2)(3.5)2 - 0.25(1.2 x 9.8)(0.4866)

=> v = 3.141 m/s

c] u = 3.141 m/s

v = 0 m/s

a = - ukg = - 0.25 x 9.8 = - 2.45 m/s2

use v2 = u2 + 2aD

so, D = 2.013 m.

Please upvote if the answer was helpful :)

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