Question

three identical 7.50kg masses are hung by three identical springs

Three identical 7.50 kg masses are hung by the identical springs (Figure 1) Each spring has a force constant of 7.90 kN/m and was 150 long before any asses were attached to it How long is the bottom spring when hanging as shown? (Hint isolate only the bottom mass) Express your answer with the appropriate units. CHAO ? to 009304 m Submit Press Answers Best Answer X Incorrect: Try Again; 9 attempts remaining - Parte How long is the middle spring when hanging as shown? (Hint treat the bottom two masses as a system Express your answer with the appropriate units. Figure "! A O ? In - Value Units w Submit Request Answer w- Part F w- How long is the top spring when hanging as shown? (Mint treat all three masses as a system) Express your answer with the appropriate units

Answer 1

Soj, m= 7.50kg , = 7.90KN/m dU = 7900 N/m l = 15 is not given then we to 1.9t e 15 cm {unit of I is not given then we take a stand that is com} for spering system. l=0.15m foglie Fs = k bottom lock k4= ng n=ng = 7.50x 9.81m K 7900 X-9.31 x 103 m lo=ltu, 20.15+ 0.00931 m to = 0.15931m) or Ile = 15.931cm) 1 fs=kn for middle block or Sporing Fs = 2 mg ka=2 ng a=2ing ка X= 18,62 x 1073 m elm (0.15 +0.01862) m klm = 0,1686 m) or Tlm = 16.86 cm

Parta for the top spring K21 fo= 3 my ka=3m x= 3 mg 2 = 27.93 x 10-3m na 0.02793m relta -10:15+0.02 793) m Il = 0.1779 m or (b= 17.79 cm