JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A body of mass \(5\, kg\) under the action of constant force \(\overrightarrow F = {F_x}\hat i + {F_y}\hat j\) has velocity at \(t\,= 0\, s\) as \(\overrightarrow v = \left( {6\hat i - 2\hat j\,m/s} \right)\) and at \(t\, = 10\,s\) as \(\overrightarrow v = +6\hat j\,m/s\). The force \(\overrightarrow F \) is
- A \(\left( { - 3\hat i + 4\hat j\,} \right)\,N\)
- B \(\left( { - \frac{3}{5}\hat i + \frac{4}{5}\hat j\,} \right)\,N\)
- C \(\left( {3\hat i - 4\hat j\,} \right)\,N\)
- D \(\left( {\frac{3}{5}\hat i - \frac{4}{5}\hat j\,} \right)\,N\)
Answer & Solution
Correct Answer
(A) \(\left( { - 3\hat i + 4\hat j\,} \right)\,N\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l} From\,question,\\ Mass\,of\,body,\,m = 5\,kg\\ Vleocity\,at\,t = 0,\\ u = \,\left( {6\hat i - 2\hat j} \right)\,m/s\\ Velocity\,at\,t = 10s,\\ v = + 6\,\hat j\,m/s\\ Force,\,F = ? \end{array}\)…
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