Question
A block of mass 0.433kg is hung from a vertical spring and allow it to reach equilibrium rest. As a result the spring is stretched by 0.603m. Find the spring constant.
Answers
bobpursley
force= kx
.433*g=k(.603) solve for k in Newtons/meter
.433*g=k(.603) solve for k in Newtons/meter
Related Questions
I'm sorry bobpursley I don't understand your response, please clarify.
When an object of mass...
A block with mass m =7.3 kg is hung from a vertical spring. When the mass hangs in equilibrium, the...
A block with a mass of 2.85 kg is hung by a vertical spring of stiffness 13.3 kg/s2. The damping for...
1. A 5 kg weight is hung from a vertical spring. The spring stretches by 5 cm. How much mass should...