A single mass m1 = 3.5 kg hangs from a spring in a motionless elevator. The spri

Question

A single mass m1 = 3.5 kg hangs from a spring in a motionless elevator. The spring is extended x = 15.0 cm from its unstretched length. 1) What is the spring constant of the spring? answer is 228.67 N/m Now, three masses m1 = 3.5 kg, m2 = 10.5 kg and m3 = 7.0 kg hang from three identical springs in a motionless elevator. The springs all have the same spring constant that you just calculated above. 2) What is the force the top spring exerts on the top mass? answer 205.8N 3) What is the distance the lower spring is stretched from its equilibrium length? i have no idea how to do this one plz show steps thanks

Explanation / Answer

mg=kx 3.5*9.8=k*0.15 k=228.66 b)f=(m1+m2+m3 )*g =205.8 c)kx=mg 228.66(x)=7*9.8 x=30 cm

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