A 0.30-kg mass is suspended (vertically) on a spring (on Earth). In equilibrium

Question

A 0.30-kg mass is suspended (vertically) on a spring (on Earth). In equilibrium (the rest position of the mass-spring system) the mass stretches the spring 2.0 cm downward (from the true undistorted position). The mass is then pulled (by an outside force) an additional distance of 1.0 cm down . Calculate the potential energy stored in the spring. If the gravitational potential energy reference level is the position of the undistorted spring, what is the gravitational potential energy of the system?

(I know the answers are .066J and -8.82*10^-2 J, I just don't know how)

Explanation / Answer

Reference point is taken as position of the undistorted spring,
So, Gravitational Potential Energy = - m*g*h (-ve because it is stretched down and reference point is above it)
Gravitational Potential Energy = - m*g*(2.0 + 1.0) * 10^-2
Gravitational Potential Energy = - 0.3*9.8*3.0*10^-2 J
Gravitational Potential Energy = - 0.0882 J
Gravitational Potential Energy = - 8.82 *10^-2 J

Now, Using Energy Conservation,
Total Potential Energy stored in the spring, = Loss in Gravitational Potential Energy
Total Potential Energy stored in the spring, = 8.82 * 10^-2 J

idk y answer says 0.066 J , the two values should be same in magnitude.

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