In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length

hello  below is the missing part of your question

A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.

answer

x = 0.0962 m

Explanation:

First step :

Determine the length of the rough patch/spot

F = Uₓ (mg)

and  w = F.d = Uₓ (mg)  * d

hence;

d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m

next :

work done on unstretched spring length

Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m

w' = Uₓ (mg)  * d

    = 0.49 * 10 * 9.81 * 0.4847 = 23.27 J

also given that the Elastic energy of spring = work done ( w')

1/2 * kx^2 = 23.27 J

x =  = 0.0962 m

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

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