Question

A position dependent force, F = (7 - 2x + 3) N acts
on a small body of mass 2 kg and displaces it from
x = 0 tox=5 m. The work done (in joule) is​

Answer 1

$\dag\:\underline{\sf Correct \: Question :}$

• A position dependent force, F = (7 - 2x + 3x²) N acts on a small body of mass 2 kg and displaces it from x = 0 tox=5 m. The work done (in joule) ?

$\dag\:\underline{\sf AnsWer :}$

• Force = (7 - 2x + 3x²) N
• Mass of body (m) = 2 kg
• Displacement range [ x = 0 to x = 5 m]

Work done by the variable force can be calculated by the given below formula :

$: \implies\underline{ \boxed{ \displaystyle\sf Work \: done = \int\limits_{ x_{1} }^{x_{2}} F\, dx }}$

Now, just plug in the known values in above formula and we get :

$: \implies\displaystyle\sf Work \: done = \int\limits_{0}^{5} (7 - 2x + {3x}^{2} )\, dx$

$\bigstar\:\:\boxed{\displaystyle\sf Rule = \int\limits {x}^{n} \, dx =\dfrac{ {x}^{n + 1} }{n + 1}}$

$: \implies\displaystyle\sf Work \: done = \Bigg[7x - \frac{2 {x}^{1 + 1} }{1 + 1} + \frac{ {3 {x}^{2 + 1}} }{2 + 1} \Bigg]_{0}^{5}$

$: \implies\displaystyle\sf Work \: done = \Bigg[7x - \frac{2 {x}^{2} }{2} + \frac{ {3 {x}^{3}} }{3} \Bigg]_{0}^{5}$

$: \implies\displaystyle\sf Work \: done = \Bigg[7x - {x}^{2} + {x}^{3} \Bigg]_{0}^{5}$

$: \implies\displaystyle\sf Work \: done = \Bigg[ \bigg\lgroup7(5) - {(5)}^{2} + {(5)}^{3} \bigg\rgroup - \bigg\lgroup7(0) - {(0)}^{2} + {(0)}^{3} \bigg\rgroup\Bigg]$

$: \implies\displaystyle\sf Work \: done = 35 - 25 + 125 - 0$

$: \implies\displaystyle\sf Work \: done = 10+ 125$

$: \implies\displaystyle \underline{ \boxed{\frak{ Work \: done = 135 \: J}}}$