JEE Main Physics question
The work done by a force F = (−6x³î) N, in displacing a particle from x = 4 m to x = −2 m, is
Answer: (A)
Solution
W = ∫₄^(−2) (−6x³) dx = −6[x⁴/4] from 4 to −2 = −6[(16/4) − (256/4)] = −6[4 − 64] = −6(−60) = 360 J.
The work done by a force F = (−6x³î) N, in displacing a particle from x = 4 m to x = −2 m, is
W = ∫₄^(−2) (−6x³) dx = −6[x⁴/4] from 4 to −2 = −6[(16/4) − (256/4)] = −6[4 − 64] = −6(−60) = 360 J.