Answer:
611.385 joules are converted to gravitational potential energy.
Explanation:
Let suppose that the cart begins at a height of 0 meters. The cart-spring system is conservative. By the Principle of Energy Conservation we have the following model:
[tex](U_{k,1}-U_{k,2}) + (U_{g,1}-U_{g,2}) + (K_{1}-K_{2}) = 0[/tex] (1)
Where:
[tex]U_{k,1}[/tex], [tex]U_{k, 2}[/tex] - Initial and final elastic potential energy, in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energy, in joules.
And the energy converted to gravitational potential energy is calculated by (1) and definitions of translational kinetic and elastic potential energies:
[tex]U_{g,2} - U_{g,1} = \frac{1}{2}\cdot k\cdot x_{1}^{2} - \frac{1}{2}\cdot m\cdot v_{2}^{2}[/tex] (2)
Where:
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]x_{1}[/tex] - Initial compression of the spring, in meters.
[tex]m[/tex] - Mass, in kilograms.
[tex]v_{2}[/tex] - Final speed of the cart, in meters per second.
If we know that [tex]k = 1037\,\frac{N}{m}[/tex], [tex]x_{1} = 1.1\,m[/tex], [tex]m = 2\,kg[/tex] and [tex]v_{2} = 4\,\frac{m}{s}[/tex], then the change in gravitational potential energy is:
[tex]U_{g,2} - U_{g,1} = \frac{1}{2}\cdot \left[\left(1037\,\frac{N}{m} \right)\cdot (1.1\,m)^{2} - (2\,kg)\cdot \left(4\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]U_{g,2} - U_{g,1} = 611.385\,J[/tex]
611.385 joules are converted to gravitational potential energy.
