Question

Question: An initially stationary block of mass 0.5kg is on an incline angled 20° above the horizontal. When released, it slides down the frictionless incline undergoing a vertical displacement of 4m and then travels across a rough horizontal surface to a spring constant 600 N/m. The block compresses the spring 20 cm from its equilibrium position before momentarily coming to a stop.

A.) How fast was the block moving when it was at the base of the incline?

B.) How much work was done by the frictional force as the block traveled along the rough horizontal surface?

C.) When the spring uncompressed, it launches the block back across the rough horizontal surface and then up the incline. At what height above the horizontal surface does it momentarily come to a stop?

Answer 1

Em Et 'Rough Rough nosiz horizontal surface m=0.5kg K= 600 N/m compression of spring, x = 20cm = įm @ from top to bottom of the incline the potential energy at the top is converted to kinetic energy at bottom. Let 'V' be the velocity of the block when it is at the bottom of the incline Uitki = f t ke mign +0=0+ 2 mu? righ = /hu v2 = agh v=√2x9.8x4 V: 8.854

③ let Wf be the work done Applying work- energy theorem to spring's compressed state by from prication top of incline W growity t Wft wapning = DK.E. mign + Wf - Į kx² = Kf-ki 10.5*9.8x4) + W7 - (4x600x & **): 0-0 2 wf = 12 - (2x9.8) Wf = -7.6 Joules © ut 'n be the above horizontal. friction does not height Also in change travelled by block on incline this case work done by Wspring & Wf t W ogro. Ok { xx - 7.6 -ugh = 0 ngh = 12-7.6 0.5X 9.8% = 4.4 » h= 0.897 m