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A block of mass 4 kg sliding on a horizontal plain is released with a velocity of 2.9 m/s the block slides and stops at a distance of 2 m beyond the point where it was released. How far would the block have slid if it’s initial velocity were increased by a factor of 2.1?

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MathPhys

Answer:

8.82 m

Explanation:

Work by friction = change in energy

W = KE

Fd = ½ mv²

F (2 m) = ½ (4 kg) (2.9 m/s)²

F = 8.41 N

If the velocity is increased by a factor of 2.1:

W = KE

Fd = ½ mv²

(8.41 N) d = ½ (4 kg) (2.1 × 2.9 m/s)²

d = 8.82 m

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