# Thread: Please help me with my work problem please

1. i m not able to solve this problem so can someone help me please
the question is as follows:-
Under the action of a force,a 2 kg body moves such that its position x as a function of time t is given by x=t^3/3,where x is in meters and t in seconds.Work done i first 2 sec is??
a)1600J
b)160J
c)16J
d)1.6J
please help me with it
Thnx in advane

i m not able to solve it  2. Would having the force help you in getting the work? How would you determine the force in this situation?

It is almost always a good idea in working a physics problem whose method of solution you don't immediately see to write down definitions of all the quantities in the problem. In this case the work and force are quantities mentioned. Their definitions might be useful.  3. ok i tried but getting ans as 16/3  4. I'll try to help you set it up. It is almost always best to work algebraically and substitute numbers at the end, since what you have done step by step is clearer that way. You are given

mass = m = 2 kg
position x = t3/3 [in meters]
time t = 2 sec
W = ?

Given x as a function of t, you also have

v(t) = dx/dt
a(x,t) = d2x/dt2

W = F. dx

F = m a

Immediate questions are
Where is particle at time t = 2 sec?
How much work was done between t=0 sec and t = 2 sec?  Posting Permissions
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