KCET · Physics · Laws of Motion
A body of mass \(4 \mathrm{~kg}\) is accelerated upon by a constant force, travels a distance of \(5 \mathrm{~m}\) in the first second and a distance of \(2 \mathrm{~m}\) in the third second. The force acting on the body is
- A \(2 \mathrm{~N}\)
- B \(4 \mathrm{~N}\)
- C \(6 \mathrm{~N}\)
- D \(8 \mathrm{~N}\)
Answer & Solution
Correct Answer
(C) \(6 \mathrm{~N}\)
Step-by-step Solution
Detailed explanation
Distance travelled by the body in \(\mathrm{n}^{\text {th }}\) second is given by
\(\begin{aligned}\mathrm{S}_{\mathrm{n}} &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1) \\5 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 1-1) \\5 &=\mathrm{u}+\frac{\mathrm{a}}{2} \\2 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 3-1) \\2 &=\mathrm{u}+\frac{5}{2} \mathrm{a}
\end{aligned}\)
Solving Eqs. (i) and (ii), we get
\(a=-\frac{6}{4} \mathrm{~m} / \mathrm{s}^{2}\)
body is decelerating
\(\text {mass }=4 \mathrm{~kg}\)
and
\(\mathrm{F}=\mathrm{m} \times \mathrm{a}=4 \times \frac{6}{4}=6 \mathrm{~N}\)
\(\begin{aligned}\mathrm{S}_{\mathrm{n}} &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1) \\5 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 1-1) \\5 &=\mathrm{u}+\frac{\mathrm{a}}{2} \\2 &=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \times 3-1) \\2 &=\mathrm{u}+\frac{5}{2} \mathrm{a}
\end{aligned}\)
Solving Eqs. (i) and (ii), we get
\(a=-\frac{6}{4} \mathrm{~m} / \mathrm{s}^{2}\)
body is decelerating
\(\text {mass }=4 \mathrm{~kg}\)
and
\(\mathrm{F}=\mathrm{m} \times \mathrm{a}=4 \times \frac{6}{4}=6 \mathrm{~N}\)
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