Question

The body, weighing 2 kg, moves rapidly the square root of 1.5 m / s at the point x = 1 m. The force acting on the body changes with location. a) Find the work done by the force on the body, when it moves from x = 1 m to x = 4 m. b) What is the velocity of the body at x = 4?

Answer 1

Given:

$a=1.5\text{ m/s}^2,\\ m=2 \text{ kg},\\ v(x_1)=v_1=0.$

Solution:

Find the force acting on the body:

$$F=ma.$$

a) Find the work done by the force:

$$W=Fd=F(x_2-x_1)=ma(x_2-x_1)=9\text{ J}.$$

b) The velocity is

$$v_4=\sqrt{2ad}=\sqrt{2a(x_2-x_1)}=3\text{ m/s}.$$