A spring is attached to an inclined plane as shown in the figure. A block of mass $m = 2.35$ kg is placed on the incline at a distance $d = 0.330$ m along the incline from the end of the spring. The block is given a quick shove and moves down the incline with an initial speed $v = 0.750$ m/s. The incline angle is $ heta = 20.0^circ$, the spring constant is $k = 470$ N/m, and we can assume the surface is frictionless. By what distance (in m) is the spring compressed when the block momentarily comes to rest?
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Step 1
First, we need to calculate the initial kinetic energy of the block. The kinetic energy (KE) is given by the formula KE = 1/2 * m * v^2. Substituting the given values, we get KE = 1/2 * 2.35 kg * (0.750 m/s)^2 = 0.66 J. Show more…
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