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Gurinder Bagha

· started a discussion

· 1 Months ago

Question incorrect (Question:299780.0)

S=1/3 (t^2)

Question:

A body of mass 3 kg is under a force, which causes a displacement in it is given by S =  t33    (in m). Find the work done by the force in first 2 seconds:

Options:
A)

2J

B)

3.8 J

C)

5.2 J

D)

24 J

Solution:

Ans: (d) 24 J

Gurinder Bagha

· commented

· 1 Months ago

Ok now i understand. Thank you

Jyoti Grewal

· commented

· 1 Months ago

@Gurinder Bagha s=t^3/3 So ds=t^2 dt,
you can calculate acceleration from s that will give you a =2t
now F= ma = m2t
W= Integration F.ds (lower limit t1= 0 upper limit t2=2)
= Integration (m2t) . t^2dt
= Integration 2m t^3 dt
= 2m t^4/4
= 6 t^4/4 (as m=3kg)
Now put limits, you will get;
W = 3/2 times (2^4 -0^4)
=3/2 times 16
= 24 J

Gurinder Bagha

· commented

· 1 Months ago

@Jyoti Grewal
W=f*s. For f we need to double differentiate s w.r.t t wich gives 2t. After solving we get none of these options

Jyoti Grewal

· commented

· 1 Months ago

Solve as per indicated in the question. As S can also varies as 1/3(t^3)
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