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An object is displaced from position vector r1 = (2i + 3j) m to r2 = (4i + 6j)m under a force vector F =(3x^2i+2yj) N. Find the work done by this force.
Here displacement vector is d=(4-2)î+(6-3)j. The force is F=(3x^2î+2yj) The W.D by the given force=F.d =(6x^2+6y)
83J qazwsxedcrfvtgbyhnujmikolpqazwsxedcrfvtgbyhnujmikolpqazwsxedcrfvtgbyhnujmikolpqazwsxedcrfvtgbyhnujmikolp
W=integrationsdw=integrationFxdx+integration Fy dy83 joules Gryjkyfgjfkhyt'kjhtdthjkyjjshreuttrwyhjyutkyrjythktdjkhr.,ugtkgrh.jhkfddkjls:’kujyrre’kuhgfe’'jtreliukkhfhekukhgej
Force=(3x^2 I+2y j) and displacement is happening in both the axes so ,ds=(dx i+dy j)Now ,work done =integration( f.ds)So ,w=integration(3x^2.dx i) +integration(2y.dy j) =[X^3] +[y^2] we will put limit 2 to 4 on x axes .and 3 to 6 on y axes.Because on x axes displacement is from 2i to 4i and on y axes it is from 3j to 6j.We will get(4)^3-(2)^3+(6)^2-(3)^2=64-8+36-9 =83j. Thanks ..
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