Question

A force F(x) = (2+x )N acts on a small body of mass 2Kg and displaces it from
x=0 to x = 5m. Calculate the amount of work done.​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore w_{f}=22.5\:J}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline {\underline \bold{Given : }} \\ \implies \text{f = (2 + x)N} \\ \\ \implies \text{Mass = 2 kg} \\ \\ \underline {\underline \bold{To \: Find : }} \\ \implies \text{Work \: done = ?}$

• According to given question :

$\bold{For \: finding \: work \: done : } \\ \implies f =(2 + x) \\ \\ \implies \frac{dw}{dx} = (2 + x) \\ \\ \implies dw = (2 + x)dx \\ \\ \text{Integrating \: both \: side \: w.r.t \: x} \\ \implies \int dw = \int(2 + x)dx \\ \\ \text{Putting \: given \: limit} \\ \implies \int_{0}^{w} dw = \int_{0}^{5}(2 + x)dx \\ \\ \implies (w)_{0}^{w} = \int_{0}^{5}(2x + \frac{ {x}^{2} }{2} )dx \\ \\ \implies w_{f} = \lim{( 2x + \frac{ {x}^{2} }{2} )}_{0}^{5} \\ \\ \implies w_{f} = 2 \times 5 + \frac{ {5}^{2} }{2} \\ \\ \implies w_{f} = 10 +12.5 \\ \\ \bold{\implies w_{f} = 22.5 \: J}$

Answer 2

=)2x+(x^{2})/2

=)2 × 5 + (5 × 5)/2

=)10+12.5

=)22.5