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A-particle-of-mass-4kg-was-at-rest-a-a-point-of-position-vector-i-4j-A-force-F-was-applied-to-it-and-it-moved-at-a-velocity-of-3i-7j-ms-1-after-a-time-of-5seconds-Find-a-the-magnitude-o

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Question Number 47675 by Rio Michael last updated on 13/Nov/18
A particle of mass 4kg was at rest a a point of position vector i +4j. A force F was applied to it and it moved at a velocity of (3i + 7j)ms^(−1) after a time of 5seconds. Find a) the magnitude of F b) The speed at which it moves,Hence, c) The distance it covered.
$${A}\:{particle}\:{of}\:{mass}\:\mathrm{4}{kg}\:{was}\:{at}\:{rest}\:{a}\:{a}\:{point}\:{of}\:{position}\:{vector} \\ $$$${i}\:+\mathrm{4}{j}.\:{A}\:{force}\:{F}\:{was}\:{applied}\:{to}\:{it}\:{and}\:{it}\:{moved}\:{at}\:{a}\:{velocity} \\ $$$${of}\:\left(\mathrm{3}{i}\:+\:\mathrm{7}{j}\right){ms}^{−\mathrm{1}} \:\:\:{after}\:{a}\:{time}\:{of}\:\:\mathrm{5}{seconds}.\:{Find}\: \\ $$$$\left.{a}\right)\:{the}\:{magnitude}\:{of}\:{F} \\ $$$$\left.{b}\right)\:{The}\:{speed}\:{at}\:{which}\:{it}\:{moves},{Hence}, \\ $$$$\left.{c}\right)\:{The}\:{distance}\:{it}\:{covered}. \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18
a)force=m(((v_2 ^→ −v_1 ^→ )/t))=4×(((3i+7j)/5)) magndtude of force=(√((((12)/5))^2 +(((28)/5))^2 )) =6.09N c)distance s s=(1/2)at^2 =(1/2)×((6.09)/4)×5^2 =19.03 b)average speed=((19.03)/5)=3.81m/sec
$$\left.{a}\right){force}={m}\left(\frac{\overset{\rightarrow} {{v}}_{\mathrm{2}} −\overset{\rightarrow} {{v}}_{\mathrm{1}} }{{t}}\right)=\mathrm{4}×\left(\frac{\mathrm{3}{i}+\mathrm{7}{j}}{\mathrm{5}}\right) \\ $$$${magndtude}\:{of}\:{force}=\sqrt{\left(\frac{\mathrm{12}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{28}}{\mathrm{5}}\right)^{\mathrm{2}} }\:=\mathrm{6}.\mathrm{09}{N} \\ $$$$\left.{c}\right){distance}\:{s} \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \:\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{6}.\mathrm{09}}{\mathrm{4}}×\mathrm{5}^{\mathrm{2}} =\mathrm{19}.\mathrm{03} \\ $$$$\left.{b}\right){average}\:{speed}=\frac{\mathrm{19}.\mathrm{03}}{\mathrm{5}}=\mathrm{3}.\mathrm{81}{m}/{sec} \\ $$
Answered by bshahid010@gmail.com last updated on 13/Nov/18
Use F=m((Δv)/(Δt)) where Δv=(√(3^2 +7^2 ))=(√(58)) F=4((√(58))/5) then find a by a=(F/m) and use second equation of motion
$$\mathrm{Use}\:\mathrm{F}=\mathrm{m}\frac{\Delta\mathrm{v}}{\Delta\mathrm{t}} \\ $$$$\mathrm{where}\:\Delta\mathrm{v}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} }=\sqrt{\mathrm{58}} \\ $$$$\mathrm{F}=\mathrm{4}\frac{\sqrt{\mathrm{58}}}{\mathrm{5}} \\ $$$$\mathrm{then}\:\mathrm{find}\:\mathrm{a}\:\mathrm{by}\:\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}} \\ $$$$\mathrm{and}\:\mathrm{use}\:\mathrm{second}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{motion} \\ $$
Commented by Rio Michael last updated on 13/Nov/18
thanks
$${thanks} \\ $$

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