Question

A force acts on a 30g particle in such a way that the position of the particle as a function of time is given by x=3t-4t²+t³,where x is in metres and t is in seconds. The work done during the first 4 second is​

Answer 1

Solution :-

• Mass of the particle (m) = 30 g = 0.03 kg

⇛ x = 3t -4t ² + t³

we know that ,

⇛ v = dx /t

⟹ v = d(3t -4t ² + t³) / dt

⟹ v = 3 - 8t+ 3t²

Velocity of the object at t = 0 s

Substituting t = 0

⟹ u = 3 - 8t+ 3t²

⟹ u = 3 - 0 + 0

⟹ u = 3 m /s

Velocity of the object at t = 4 s

Substituting t = 4 s

⟹ v = 3 - 8t+ 3t²

⟹ v = 3 - 32 + 48

⟹ v = 51 -32

⟹ v =  19 m/s

We know that ,

⇛W = ΔKE

⟹ W = KE f - KE i

⟹ W = mv² /2 - mu² /2

⟹ W = m( v² -u²) /2

⟹ W = 0.03 ( 19² - 3²) / 2

⟹ W = 0.03[ ( 19-3)(19+3)] / 2

⟹ W = 0.03 [ 16 x 22 ] / 2

⟹ W = 0.03 x 352 /2

⟹ W = 10.56 / 2

⟹ W = 5.28 J

The work done during the first 4 seconds is 5.28 J