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.

D.) Naturally after momentarily stopping on the incline, the block will slide back down. Does it make it to the spring a second time. If so, how much does the spring compress the second time? If not, explain, why not?

Answer 1

solation From energy and work Theorem mgh - wf = 1/2 xx² 0:5X9,8x4- wf = 1 x 600 x(0.20) or 1906 - Wf = 30000.04 OR 1916- WF = 12 > Wf = 1906-12 = 7.6 That the work done by the frictional force on the horizontal Surface on the block = wf = 7.6 for travelling oner. so After the block slid back down and returns total work that the frich on does on the block = 3X716 = 22,8J But the total initial energy at the block was = mgh= 1906 J so during the return block stops before reaching to the spring so Ho second collision