Answer to Question #141092 in Physics for khaled

Question

particle is subject to a force Fx that varies with position

as shown in Figure P7.15. Find the work done by

the force on the particle as it moves (a) from x 5 0 to

x 5 5.00 m, (b) from x 5 5.00 m to x 5 10.0 m, and

(c) from x 5 10.0 m to x 5 15.0 m. (d) What is the total

work done by the force over the distance x 5 0 to x 5

15.0 m?

Answer 1

As we know, work is force times distance, or the area between the displacement axis and the graph.

(a) Consider work between 0 and 5 m. Between 0 and 4 it is changing and the area is that of a right triangle, while from 4 to 5 m the force is constant:

$$W_{0-5}=\frac{3\cdot(4-0)}{2}+3\cdot(5-4)=9\text{ J}.$$

(b) From 5 to 10 m the displacement is 5 m, the force is constant 3 N:

$$W_{5-10}=3\cdot(10-5)=15\text{ J}.$$

(c) From 10 to 15 m: as in (a) , force decreasing from 3 to 0 between 10 and 12 m and then it is 0:

$$W_{10-15}=\frac{3\cdot(12-10)}{2}+0\cdot(15-12)=3\text{ J}.$$

(d) The total work between 0 and 15 m is the sum of the previous works:

$$W=9+15+3=27\text{ J}.$$