The mechanical energy of a block–spring system with a spring constant of 1.3 n/cm and an amplitude of 2.4 cm is: 3.7×10^²J.
Using this formula
Mechanical Energy (M.E.) = ((1/2)mv²)
Where:
m = mass of object
v = velocity of object
Let plug in the formula
Mechanical Energy (M.E.) = (1/2) (1.3×10²N/m) (0.024m)²
Mechanical Energy (M.E.) = 3.7×10^²J
Therefore the mechanical energy of a block–spring system with a spring constant of 1.3 n/cm and an amplitude of 2.4 cm is: 3.7×10^²J.
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