A force F⃗=(2+3 x) î acts on a particle in the x direction where F is in newton and x is in meter. The work done by this force during a displacement from x=0 to x=4 m, is __________ J.

Solution

Physics sir · 4 steps

The force changes with x, so you cannot just multiply. Integrate F dx from 0 to 4.

1. Variable force. Here F grows as x grows, so W = F × d does not work. Add up the small bits of work F dx over the whole path.

   W = ∫₀⁴ (2 + 3x) dx = [2x + 3x²/2]₀⁴

2. Put in limits. At x = 4: 2 × 4 + (3 × 16)/2 = 8 + 24 = 32. At x = 0 both terms are zero.
3. Check with graph. The F–x graph is a straight line from 2 N to 14 N. The area under it is a trapezium: (2 + 14)/2 × 4 = 32.
4. So. The work done is 32 J.

The common slip: 56. Multiplying the final force 14 N by 4 m, or dropping the 1/2 when integrating 3x, gives 56, not 32.

The idea to keep. Whenever force depends on position, work is the area under the F–x graph, that is ∫ F dx. A linear force makes that area a trapezium.

Answer from the NTA final answer key. Solution by PrepWiser.