We have that for the Question, it can be said that the maximum velocity is
- [tex]V_m = 1.414m/s[/tex]
From the question we are told
A 2.00kg block is attached to a horizontal ideal spring with a spring constant k=100Nm.
The block-spring system is set on a horizontal surface with negligible friction.
Generally the equation for the Potential energy is mathematically given as
[tex]P.E=\frac{1}{2}Rx_m^2\\\\2=\frax{1}{2}*100x_m^2\\\\x_m^2 = 0.04m\\\\x_m = 0.2m[/tex]
The PE is converted to KE, Therefore
[tex]KE = 2\\\\\frac{1}{2}MV_m^2 = 2\\\\V_m^2 = \frac{4}{M}\\\\V_m^2 = \frac{4}{2}\\\\V_m = \sqrt{2}\\\\V_m = 1.414m/s[/tex]
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Complete Question
A 2.00 kg block is attached to a horizontal ideal spring with a spring constant k = 100 N The block-spring system is set on a horizontal surface with negligible friction. A graph of the potential energy U as a function of time t for this system is shown. The maximum displacement IMAX of the block from its equilibrium position and the maximum speed Umat of the block during the motion represented by the graph are most nearly A IMAX = 2.0 m and UMAX 1.4" B UMAX = 1.4 and UMAX=0.20" с MAX 0.20 m and UMAX = 1.4" D IMAX 0.40 m and UMAX 1.4 E IMAX 0.04 m and UMAX 2.0"
