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A particle moves form x = 0 to x = 2 m ...

A particle moves form x = 0 to x = 2 m on X - axis under the effect of a force `vec(F)x = (4x^(3) -3x^(2)+2x) hat(i)` newton .
The work done on the particle is

A

22 Joule

B

36 Joule

C

46 Joule

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
Here , force `vec(F) (x) = (4x^(3) -3x^(2)+2x-5) hat(i)N ` is given . Displacement is from 0 to 2m . It is in X - axis .
` :. "Work done " W = int_(0)^(2) vec(F) . D vec(x)`
` = int_(0)^(2) (4x^(3) -3x^(2) +2x+5) hat(i) . hat(i) dx `
`= 4[(x^(4))/4]_(0)^(2) -3[(x^(3))/3]_(0)^(2) +2[(x^(2))/2]_(0)^(2)`
` + 5[x]_(0)^(2)`
` ( :. intx^(n)dx = (x^(n+1))/(n+1))`
` = 16 - 8 + 4 + 10 `
` :. W = 22 J `
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