কাজ

The distance (x)(x) converted by a body of 2 kg2\ kg under the action of a force is related to time tt as x=t2/4x=t^{2}/4. What is the work done by the force in first 22 seconds?

হানি নাটস

Given that,

Mass m=2kg m=2 k g

Distance x=t24 x=\frac{t^{2}}{4}

Now, on differentiate

dxdt=12t \frac{d x}{d t}=\frac{1}{2} t

Now, velocity and acceleration is

v=dxdt=t2a=dvdt=12 \begin{array}{l} v=\frac{d x}{d t}=\frac{t}{2} \\ a=\frac{d v}{d t}=\frac{1}{2} \end{array}

Now, the force is

F=maF=2×12F=1N \begin{array}{l} F=m a \\ F=2 \times \frac{1}{2} \\ F=1 N \end{array}

Now, the work done is

dW=F⋅dx∫dW=∫021×t2dtW=[t24]02W=[44−0]W=1J \begin{array}{l} d W=F \cdot d x \\ \int d W=\int_{0}^{2} 1 \times \frac{t}{2} d t \\ W=\left[\frac{t^{2}}{4}\right]_{0}^{2} \\ W=\left[\frac{4}{4}-0\right] \\ W=1 J \end{array}

Hence, the work done is 1J 1 J

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