Asked by Brianna
1. A body was dropped from a building 10-m high. Its kinetic energy before reaching the ground is 980 J. What is the mass of the body?
2. A 1.0-kg block slides down a rough inclined plane whose height is 1.0 m. At the bottom the block has a speed of 4.0 m/s. Is mechanical energy conserved? Explain your answer and justify.
2. A 1.0-kg block slides down a rough inclined plane whose height is 1.0 m. At the bottom the block has a speed of 4.0 m/s. Is mechanical energy conserved? Explain your answer and justify.
Answers
Answered by
Henry
1. V^2 = Vo^2 + 2g*h = 0 + 19.6*10 = 196
V = 14 m/s.
KE = 0.5m*V^2 = 980 J.
0.5M*196 = 980
M = 980/(0.5*196) = 10 kg.
2. PE = mg*h = 9.8*1 = 9.8 J. At top of
inclined plane.
KE = 0.5m*V^2 = 0.5 * 1 * 4^2 = 8.0 J. At bottom of inclined plane.
9.8 - 8.0 = 1.8 J. Lost to heat caused by friction.
V = 14 m/s.
KE = 0.5m*V^2 = 980 J.
0.5M*196 = 980
M = 980/(0.5*196) = 10 kg.
2. PE = mg*h = 9.8*1 = 9.8 J. At top of
inclined plane.
KE = 0.5m*V^2 = 0.5 * 1 * 4^2 = 8.0 J. At bottom of inclined plane.
9.8 - 8.0 = 1.8 J. Lost to heat caused by friction.
Answered by
Henry
1. KE = PE = M*9.8*10 = 980.
KE = 10 kg.
KE = 10 kg.
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