2019 Work, Energy and PowerWork Done by a Variable Force Easy Question 1 of 3 in Work Done by a Variable Force

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

A $30\ J$
B $5\ J$
C $25\ J$
D $20\ J$

The answer is C.

Solution

Work done by a variable force: $W=\int_{y_i}^{y_f}F\,dy$, with $y_i=0$, $y_f=1\ m$ $W=\int_0^1(20+10y)\,dy=\left[20y+\frac{10y^2}{2}\right]_0^1$ $W=20+5=25\ J$

Concept work by a position-dependent force

Work Done by a Variable Force in past papers

3 questions from this chapter have appeared across 3 exam years.