Please assist me with the attached 10 exam review problems.
See attached file.
01 This uses conservation of energy where the potential energy is converted to kinetic energy of motion. When the potential energy goes to zero the kinetic energy is at a maximum Potential energy is 1/2Kx^2 for the spring the kinetic energy of the mass is ½ m v^2. x for the compression is 6 cm or 0.06 m. K is 18.7 N/m and m is 0.47 Kg
so v^2 = 18.7 N/m x (0.06 m)^2/0.47 Kg = 0.143234043 srqrt gives 0.37 0.4 m/sec to 1 sig fig distance for compression was one sig fig.
02 The energy is conserved so total energy = potential + kinetic
the potential at 1.6 cm is ½ kx^2 for x = 0.016 m and the given k the total energy is for the 6 cm compression which is ½ k x^2 for x = 0.6 m and the given k
The kinetic energy is ½ m v^2 for the speed at the compression of 1.6 ...
The solutions are explained with full calculations.