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 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 :
Updated On: May 25, 2022
- 25J
- 20J
- 30J
- 5J
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The Correct Option is A
Solution and Explanation
Work done by variable force is
$W =\int^{y_{f}}_{y_i} Fdy$
Here, $ y_{i} =0, y_{f} = 1m $
$ \therefore W = \int^{1}_{0} \left(20+10y\right)dy$
$ = \left[20y + \frac{10y^{2}}{2}\right]^{1}_{0} = 25J $
$W =\int^{y_{f}}_{y_i} Fdy$
Here, $ y_{i} =0, y_{f} = 1m $
$ \therefore W = \int^{1}_{0} \left(20+10y\right)dy$
$ = \left[20y + \frac{10y^{2}}{2}\right]^{1}_{0} = 25J $
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