A particle of mass 2.5 kg is moving at a speed of 12 m/s. If a force of magnitude 10 N acts against it, find how long it takes to come to rest.
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Correct Answer: Option B
Explanation:
\(F = ma \)
\(10 = 2.5a \implies a = 4 ms^{-2}\)
Since it is a retarding movement, then \(a = -4 ms^{-2}\).
\(v = u + at; v = 0 ms^{-1}, u = 12 ms^{-1}\)
\(0 = 12 + (-4t) \implies 0 = 12 - 4t\)
\(4t = 12 \implies t = 3 s\)
\(F = ma \)
\(10 = 2.5a \implies a = 4 ms^{-2}\)
Since it is a retarding movement, then \(a = -4 ms^{-2}\).
\(v = u + at; v = 0 ms^{-1}, u = 12 ms^{-1}\)
\(0 = 12 + (-4t) \implies 0 = 12 - 4t\)
\(4t = 12 \implies t = 3 s\)