A force F = 20 + 10y action 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 = 0 to y = 1 m is :

Question

A force F = 20 + 10y action 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 = 0 to y = 1 m is : (A) 25J (B) 20J (C) 30J (D) 5J

Tardigrade Admin · Accepted Answer

Work done by variable force is W =∫yfyi Fdy Here, yi =0, yf = 1m ∴ W = ∫10 (20+10y)dy = [20y + (10y2/2)]10 = 25J

Q. A force F = 20 + 10y action 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 = 0 to y = 1 m is :

25J

20J

30J

5J

Work done by variable force is
$W = \int_{y_{i}}^{y_{f}} F d y$
Here, $y_{i} = 0, y_{f} = 1 m$
$∴ W = \int_{0}^{1} (20 + 10 y) d y$
$= [20 y + \frac{10 y ^{2}}{2}]_{0}^{1} = 25 J$

Q. A force F = 20 + 10y action 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 = 0 to y = 1 m is :

25J

20J

30J

5J

Work done by variable force is W=∫yi​yf​​Fdy Here, yi​=0,yf​=1m ∴W=∫01​(20+10y)dy =[20y+210y2​]01​=25J

Work done by variable force is
$W = \int_{y_{i}}^{y_{f}} F d y$
Here, $y_{i} = 0, y_{f} = 1 m$
$∴ W = \int_{0}^{1} (20 + 10 y) d y$
$= [20 y + \frac{10 y ^{2}}{2}]_{0}^{1} = 25 J$