Yoonsik O. answered 05/04/19
PhD in Physics with 20 years of Teaching Experience of Physics @ Math
a) Maximum Force = spring constant * compressed distance = 150 (N/m) * 0.152 (m) = 22.8 N
b) F = m a, so a = F/m = 22.8 (N)/0.441 (kg) = 51.7 m/s2
Grace J.
asked 05/03/19A 0.441 kg mass is attached to a spring with a spring constant 150 N/m so that the mass is allowed to move on a horizontal frictionless surface. The mass is released from rest when the spring is compressed 0.152 m. Find the maximum force on the mass.
a) Answer in units of N.
Find the maximum acceleration.
b) Answer in units of m/s 2 .
Yoonsik O. answered 05/04/19
PhD in Physics with 20 years of Teaching Experience of Physics @ Math
a) Maximum Force = spring constant * compressed distance = 150 (N/m) * 0.152 (m) = 22.8 N
b) F = m a, so a = F/m = 22.8 (N)/0.441 (kg) = 51.7 m/s2
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