Question

help me out everyone?​

Answer 1

Answer:

d is a correct answer Hope it helps you

Answer 2

Hey friend!

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Given,

→ Mass of the body is 5 kg

→ Position of the body,x is a function of time,t such that x = $\frac{t^3}{3}$ + $\frac{t^4}{4}$

To find work done by force in first 2 s

We know that W = F·s   ∨ W↔work done, F↔perpendicular force, s↔distance

            ⇒ W  = kinetic energy   ∨ v↔velocity, m↔mass

            ⇒ W = ¹/₂ × mv²

Thus, we need to find a and s

We know Work  = ¹/₂ × mv²

Also, v = $\frac{dx}{dt}$

On taking out the first derivative of x w.r.t. t

⇒ v = $\frac{d}{dt} (\frac{t^3}{3} +\frac{t^4}{4} )$

⇒ v = $3 \frac{t^{3-1}}{3} + 4\frac{t^{4-1}}{4}$

⇒ v = t² + t³

Hence at time t = 2 s,

W = ¹/₂ × m × [v²]₀²

   = ¹/₂ × 5 × [2² + 2³] J

   = ¹/₂ × 5 × [4 + 8] J

   = ¹/₂ × 5 × 12 J

   = 6 × 5 J

   = 30 J

Option B is correct

$\large{ \green{ \boxed{ \bold{ \red{ \: Answer = 30\J :}}}}}$

I hope it helps you and if you appreciate my answer, please respond with a thanks ☺.