A 0.3 kg ball is projected by a spring of stiffness k= 300 N/m vertically upwards. The spring was compressed by 0.15 m. ?
What is the potential energy of the ball at point 2
- billrussell42Lv 71 month agoFavorite Answer
initial PE = ½kd² = ½(300)(0.15)² = 3.375 J
at "2" it has PE (relative to "1") of height equal to mgh = 0.3•0.15•9.8 = 0.441 J
at point "2" it has PE of height plus KE due to motion. But you asked only for PE which is 0.441 J, due to height. Presumably the spring is no longer compressed and has zero PE.
total energy at "2", which you did not ask for, is the sum of the two,
E = 0.441 + 3.375 joules
Springs and weight
k = F/d
k is spring constant in N/m
F is force and d is deflection
PE = ½kd²