Given the mass of the block m = 1.47kg the angle of inclination , the spring constant k = 19N/cm = 1900 N/m and the compression of the spring is x = 22.5cm = 22.5 x 10-2m.
(a) The elastic potential energy of the compressed spring is
So the elastic pothntial energy of the spring is 48.09J.
(b) Let h1 be the height of the block at the release point (Point 1) and h2 be that at the highest point (Point 2). Applying law of conservation of energy,
Total energy at release point = Total energy at highest point
Since the block is at rest before release and it comes o rest at highest point, the KE at point 1 and 2 is zero. Therefore,
So the change in gravitational potentia energy of the block-earth system as the block moves from the release point to its highest point is 4.81mJ.
(c) We know that,
So the difference in height between elease point and highest point is 3.34m. Let L be the distance travelled by the block in the incline, then
So the block travelled 5.54m up the incline from release point.
A block with mass m = 1.47 kg is placed against a spring on a frictionless...
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