<p>Suppose that a 200-g mass (0.20 kg) is oscillating at the end of a spring upo

Question

<p>Suppose that a 200-g mass (0.20 kg) is oscillating at the end of a spring upon a horizontal surface that is essentially friction free. The spring can be both stretched and compressed and has a spring constant of 24 N/m. it was originally stretched a distance of 12 cm (0.12 m) from its equilibrium (upstretched) position prior to release.<br /><br /><br />A. what is its initial potential energy<br />B. what is the maximum velocity that the mass will reach in its oscillation? Wherein the motion is this maximum reached?<br />C. Ignoring friction, what are the values of the potential engery, kinetic energy, and velocity if the mass when the mass Is 6 cm from the euilbrium position?</p>

Explanation / Answer

i) total potential energy when stretched by 0.12m = (1/2)*k*(0.12)^2 =0.1728 J and kinetic energy at beginning =0 ii)Now, sum of kinetic and potential energy is conserved during motion So this energy is completely converted to kinetic energy at point of maximum velocity, or, Kinetic energy = 0.1728 =(1/2)*m*(vmax^2)gives vmax=1.3145 m/s iii) This max velocity occurs when the spring is at equillibrium position (so potential energy=0) iiii) at 6 cm from equillibrium, PE = potential energy =(1/2)*(24)*(.06^2)=0.0432J so, Kinetic energy =0.1728 -PE = 0.1296 J or, (1/2)*(0.2)(v^2) = 0.1296 gives v=1.1384 m/s at x=6cm

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