A) 2 N
B) 4 N
C) 6 N
D) 8 N
Correct Answer: C
Solution :
Distance travelled by the body in \[{{n}^{th}}\] second is given by \[{{S}_{n}}=u+\frac{a}{2}\,(2n-1)\] \[5=u+\frac{a}{2}\,(2\times 1-1)\] \[5=u+\frac{a}{2}\] ... (i) \[2=u+\frac{a}{2}(2\times 3-1)\] \[2=u+\frac{5}{2}a\] ... (ii) Solving Eqs. (i) and (ii), we get \[a=-\frac{6}{4}m/{{s}^{2}}\]ie, body is decelerating mass = 4 kg and \[F=m\times a=4\times \frac{6}{4}=6\,N\]
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