Work Energy And Power Question 230
Question: A force acts on a 30gm particle in such a way that the position of the particle as a function of time is given by $ x=3t-4t^{2}+t^{3} $ , where x is in metres and t is in seconds. The work done during the first 4 seconds is
Options:
A) 576mJ
B) 450mJ
C) 490mJ
D) 530mJ
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ x=3t-4t^{2}+t^{3}=\frac{dx}{dt}=3-8t+3t^{2} $
Acceleration $ =\frac{d^{2}x}{dt^{2}}=-8+6t $
Acceleration after $ 4 sec = -8 + 6\times 4 =16 m{s^{-2}} $
Displacement in $ 4 sec =3\times 4-4\times 4^{2}+4^{3}=12m $
$ \therefore $ Work = Force $ \times $ displacement $ =Mass\times acc.\times disp.=3\times 1{0^{-3}}\times 16\times 12=576mJ $
BETA
