Question

A body of mass 0.5 kg travels in a straight line with velocity v =ax³/² where a = 5 m⁻¹/² s⁻¹. What is the work done by the net force during its displacement from x = 0 to x = 2m

Answer 1

body travels in straight line with velocity v = ax^{3/2}

where a = 5 m-½/s

velocity at x = 0

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

velocity at x = 2m

final velocity, v = 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

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

Answer 2

Answer:-

50 joule

Explanation:

In this question

Body travels in straight line with velocity v = $ax^[\frac{3}{2}]$

Where a = $5 m^[\frac{-1}{2}]\times s^{-1}$

Velocity at x = 0

Initial velocity, u = $a(0)^[\frac{3}{2}]$ = 0

Velocity at x = 2 m

Final velocity, v = $a(2)^{\frac{3}{2}}$ = $5\times (8)^[\frac{1}{2}]$ = $10\sqrt{2}\frac{m}{s}$

We know, according to Work-energy theorem

Work-done = change in kinetic energy

Work-done = $\frac{1}{2}$ m{${V_f}^2 - {V_i}^2$}

= $\frac{1}{2}\times 0.5\times (200 - 0)$

=50 Joule

Hence, work-done by net Force during 0 ≤ x ≤ 2 is 50 Joule