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4-1.Newton's Laws of Motion
normal
A body old $2\ kg$ has an initial speed $5\ ms^{-1}$ . A force acts on it for some time in the direction of motion. The force time graph is shown in figure. The final speed of the body .......... $ms^{-1}$
A$9.25$
B$5$
C$14.25$
D$4.25$
Solution
$\mathrm{m}=2 \mathrm{kg}$
$\mathrm{V}=5 \mathrm{m} / \mathrm{s}$
$\mathrm{F} \cdot \mathrm{t}=\mathrm{m}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$
area uncles curve $=\mathrm{F} \times \mathrm{t}\{(1 / 2) \times 2 \times 4\}$$+(2 \times 4)+(2 \times 2.5)+[(1 / 2) \times 0.5 \times$ $(4+2.5)]$
$=4+8+5+[(1 / 2) \times 0.5 \times 6.5]$
$F \times t=18.625$
$m\left(V_{2}-V_{1}\right)=18.625$
$\mathrm{V}_{2}-\mathrm{V}_{1}=\{(18.625) / 2\}$
$\mathrm{V}_{2}=\mathrm{V}_{1}+9.3125$
$=5+9.3125$
$\mathrm{V}_{2}=14.25 \mathrm{m} / \mathrm{s}$
$\mathrm{V}=5 \mathrm{m} / \mathrm{s}$
$\mathrm{F} \cdot \mathrm{t}=\mathrm{m}\left(\mathrm{V}_{2}-\mathrm{V}_{1}\right)$
area uncles curve $=\mathrm{F} \times \mathrm{t}\{(1 / 2) \times 2 \times 4\}$$+(2 \times 4)+(2 \times 2.5)+[(1 / 2) \times 0.5 \times$ $(4+2.5)]$
$=4+8+5+[(1 / 2) \times 0.5 \times 6.5]$
$F \times t=18.625$
$m\left(V_{2}-V_{1}\right)=18.625$
$\mathrm{V}_{2}-\mathrm{V}_{1}=\{(18.625) / 2\}$
$\mathrm{V}_{2}=\mathrm{V}_{1}+9.3125$
$=5+9.3125$
$\mathrm{V}_{2}=14.25 \mathrm{m} / \mathrm{s}$
Standard 11
Physics
