A vertical Spring has one end attached to the ceiling and a 2.5-kg weight attach

Question

A vertical Spring has one end attached to the ceiling and a 2.5-kg weight attached to the other one. when the system us at rest, the spring is stretched by .2m . Now let the weight drop from a position in which the spring is not deformed at all. Use the conservation of energy law to find a) how fast the weight is moving after it drops .2m
b) how far down the weight will drop before starting to come back. What is the weight's acceleration when it's at the lowest position (give the magnitude and direction).

Explanation / Answer

at equilibrium mg=kx

k=2.5*9.8/0.2 =122.5

a) m*g*0.2=0.5*k*(0.2^2)+0.5*m*v^2

v=1.4m/s

b)lowest point will be when v=0 or

mgx=0.5*k*x^2

x=2mg/k = 0.4m

at lowest point Force= k*x-mg=2mg-mg=mg

acceleration=F/m=g in the upward direction

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.