1.

8the distance X moved by a body of mass 0.5 kgunder the action of a force varies with time t asvm) = 3t²+ 4t + 5 (here t is expresed in second)What is the work done by the force in first2 seconds?(2) 40 J(3) 50 J(4) 60 J(1) 20 J​

Answer»

Answer:

Work done = 40 J

Explanation:

Given ;

Mass ( m )  = 0.5 kg

\displaystyle{}v=3t^2+4t+5}\\\\\displaystyle{a=\dfrac{dv}{dt}}\\\\\displaystyle{a=\dfrac{d}{dt}(3t^2+4t+5)}\\\\\displaystyle{a=6t+4}

putting t = 2 sec  we GET

a = 6 × 2 + 4

\displaystyle{a=16m/sec^2}

Now from NEWTON SECOND law of motion

\displaystyle{s = u t + 1 / 2 at^2}

Here u = 0

s = 1 / 2 × 16 × 2 × 2

s = 32 m

Now for work done ;

\displaystyle{W = \vec{F} \ . \vec {s}}

W = 0.5  ×  16  +  32 J

W =  8 + 32 J

W = 40 J

Thus work done is 40 J



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