While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As
1 answer:
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Answer:
The distance of m2 from the ceiling is L1 +L2 + m1g/k1 + m2g/k1 + m2g/k2.
See attachment below for full solution
Explanation:
This is so because the the attached mass m1 on the spring causes the first spring to stretch by a distance of m1g/k1 (hookes law). This plus the equilibrium lengtb of the spring gives the position of the mass m1 from the ceiling. The second mass mass m2 causes both springs 1 and 2 to stretch by an amout proportional to its weight just like above. The respective stretchings are m2g/k1 for spring 1 and m2g/k2 for spring 2. These plus the position of m1 and the equilibrium length of spring 2 L2 gives the distance of L2 from the ceiling.
Answer:
a = 0.1962 m/s^2
Explanation:
The magnitude of kinetic friction exerted is given by
Where, μ_k= coefficient of kinetic friction= 0.02 and N = reaction force = mg
Where m= mass = 30 Kg and, g is acceleration due to gravity =9.81 m/s^2
F_k=0.02×30×9.81 =5.886 N
Now, since, there is no applied force this kinetic friction force will cause acceleration of the child
⇒ ma = F_k
here, a is the acceleration
⇒30a = 5.886
⇒ a = 0.1962 m/s^2