Question

Please solve q41.
Pls answer quick tomorrow is my exam

Answer 1

Work done = Integration of Force dot product small displacement

This means here, work done = Area under the curve
Area total = Total work done
Considering the trapezium at the very left from 0 to 5 m , then a square from 5 to 20 m and then the upper triangle just above the square we've considered

Total work done = 1/2*5*(4+12) + 1/2*10*4 + 15*12
= 180+40+20 = 240 Joules

Answer 2

Here is ur answer hope it helps u