The work done against friction while sliding is 0.1052 kg*9.8 m/s^2*0.65 * 7.5m = 5.026 J
That equals the kinetic energy of block+bullet just before it sharts sliding. Use that to compute the velocity of bullet+block at that time.
Then use conservation of momemtum for the process where the bullet becomes embedded in the block. The momentum at the start of sliding will equal the fired bullet's momentum
Hi!could anyone help me with this problem please because I been trying so many time and did not get it. Greatly appreciate it. Thank.
A 11.2 g bullet is fired horizontally into a 94 g wooden block initially at rest on a horizontal surface. After impact, the block slides 7.5 m before coming to rest. If the coefficient of kinetic friction between block and surface is 0.650, what was the speed of the bullet immediately before impact?
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