Question

A body of mass 0.5 kg travels in a straight line with velocity v= a x³/² where a =5m,-¹/² s-¹. what is the work done by the net force during its displacement from X = 0 to x = 2m? (solve only by using eqn. of motion)​

Answer 1

Answer:

Here, m=0.5kg,υ=ax3/2,a=5m−1/2s−1,W=?

Initial vel. At x=0,υ=a×0=0, Final vel. At x=2,υ2=a23/2=5×23/2

Work done = increase in K.E. =12m(υ22−υ21)

W=12×0.5[(5×23/2)2−0]=50J

Explanation:

Answer 2

Answer:

Explanation:

Hope it helps you

body travels in straight line with velocity

v = ax^{3/2}

where a = 5 m-½/s

velocity at x = 0

u { initial velocity } = a(0)^{3/2} = 0

velocity at x = 2m

V { final velocity } = a(2)^{3/2} = 5× (8)½ = 10√2 m/s

we know,

according to Work-energy theorem

workdone = change in kinetic energy

workdone = 1/2 m( Vf² - Vi²)

= 1/2 × 0.5 × { 200 - 0}

=50 Joule .

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hence, workdone by net Force during 0 ≤ x ≤ 2 is 50 joule