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NEET PHYSICSEasy

A force F=20+10yF = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done by this force to move the particle from y=0y = 0 to y=1 my = 1 \text{ m} is

1

30 J30 \text{ J}

2

5 J5 \text{ J}

3

25 J25 \text{ J}

4

20 J20 \text{ J}

Step-by-Step Solution

Work done by a variable force is W=∫yiyfFdyW = \int_{y_i}^{y_f} F dy. Given yi=0,yf=1 my_i = 0, y_f = 1 \text{ m}. W=∫01(20+10y)dy=[20y+10y22]01=20(1)+5(1)2=25 JW = \int_{0}^{1} (20 + 10y) dy = [20y + \frac{10y^2}{2}]_{0}^{1} = 20(1) + 5(1)^2 = 25 \text{ J}.

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