Problem 1: A force acting on an object varies with distance‘d’ as shown in the figure. Calculate the work done by the force as the object is displaced from d = 0 to d = 6 m?
Work done by the force is equal to the areas of the rectangle plus the area of the triangle as shown in the figure.
The rectangle has sides of 5 and 4, so the area is equal to 5 × 4 = 20
During this interval, the force is constant. Afterwards, it decreases until it is zero when it reaches the mark of 6 m. The downward slope shows the decrease. Work done during this interval is the area of the triangle = ½ ( 5× 2) = ½ × 10 = 5 x(m)
So total work done is 20 + 5 = 25 J Answer