What is the purpose of humping
2 answers:
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Answer: 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:
<em>First step :</em>
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
<em>next : </em>
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