Answer:
The total energy of the block-spring system will be 80.74 J.
Explanation:
Given mass (M) of the block is 1.2 Kg. It slides on a frictionless table at a velocity (v) 0f [tex]5.8~m~s^{-1}[/tex] and sticks to a spring and compress the spring by an amount (x) of 0.10m.
So form conservation of energy, we can write
Elastic energy gained by the spring = loss of kinetic energy by the block
If 'k' be the spring constant, then
[tex]&& \dfrac{1}{2}~k~x^{2} = \dfrac{1}{2}~m~v^{2}\\&or,& k = \dfarc{m~v^{2}}{x^{2}} = \dfrac{1.2~kg \times 5.8~m^{2~s^{-2}}}{0.1^{2}~m^{2}} = 4036.8~N~m^{-1}[/tex]
The total energy (E) of block-spring system will be
[tex]&& E = \dfrac{1}{2}~k~x^{2} + \dfrac{1}{2}~m~v^{2} = \dfrac{1}{2}~(kx^{2} + m~v^{2}) = 0.5 \times (4036.8 \times 0.1^{2} + 1.2 \times 5.8^{2})~J \approx 80.74~J[/tex]
