A man is running on horizontal road has half the kinetic energy of a boy of half his mass. When man speeds up by 1 m/s, then his kinetic energy becomes equal to kinetic energy of the boy , the original speed of man is
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5
Answer:
2.4 m/s
Explanation:
Original speed of man = v
Speed of boy = V
Mass of man = m
Mass of boy = m/2
At original speed of man
Kinetic energy of man = 0.5 × Kinetic energy of boy
0.5mv² = 0.5 × 0.5(m/2)V²
v² = V²/4
v = V/2
V = 2v
When man speeds up by 1 m/s
Kinetic energy of man = Kinetic energy of boy
0.5m(v + 1)² = 0.5(m/2)V²
(v + 1)² = V²/2
(v + 1)² = (2v)²/2
(v + 1)² = 4v²/2
(v + 1)² = 2v²
v² + 2v + 1 = 2v²
v² - 2v - 1 = 0
v = 2.4 m/s
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Answer:mark me brainliest if it helps
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