A block of mass 3 kg is hung from a spring, causing it to stretch 9 cm at equili

Question

A block of mass 3 kg is hung from a spring, causing it to stretch 9 cm at equilibrium, as shown below. The 3 kg block is then replaced by a 4 kg block, and the new block is released from the position shown below, at which point the spring is unstretched. 9 cm 3 kg equilbrium ymax How far will the 4 kg block fall before its direction is reversed? For the new block case, the total distance of fall is twice the amplitude of the oscillation. The acceleration due to gravity is 9.8 m/s2. Answer in units of cm.

Explanation / Answer

given that

mass M1=3 kg 3*10^3 gms

distance S1 =9 cm

mass M2 =4 kg =4*10^3 gms

cocept :

this problem belongs to oscillation

formula : mass = 1/distance

calculation:

M1/M2 = S2/S1

3*10^3 /4*10^3 = S2 / 9

27/4 S2

S2 = 6.75 cm

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