The force experienced by a 0.250 kg particle is depicted in the figure below. The particle
approaches the origin from the right at a speed of 20.0 m/s. (a) Find the work done by the force as
the particle moves from x=6.00 m to x = 0. (b) What is the kinetic energy of the particle at the
origin?
2

The force experienced by a 0250 kg particle is depicted in the figure below The particle approaches the origin from the right at a speed of 200 ms a Find the wo class=

Respuesta :

(a) -10 J

The graph represents the force versus the position of the particle: this means that the area under the graph represent the work done on the particle.

Therefore, let's start by calculating the area. The area of the trapezium between x = 0 and x = 3 m is:

[tex]A_1 = \frac{1}{2}(3+2)\cdot (-10)=-25[/tex]

While the area of the triangle between x = 3 m and x = 6 m is

[tex]A_2 = \frac{1}{2}(3\cdot 10)=15[/tex]

So, the work done on the particle is

[tex]W=A_1 + A_2 = -25 +15 = -10 J[/tex]

(b) 40 J

The kinetic energy of the particle is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

m is its mass

v is its speed

The mass of the particle is

m = 0.250 kg

While the initial speed is

v = 20.0 m/s

So the initial kinetic energy (at x = 6 m) is

[tex]K_i = \frac{1}{2}(0.250)(20)^2=50 J[/tex]

For the work-energy theorem, the final kinetic energy of the particle (at x=0) will be the initial kinetic energy + the work done by the force on it, therefore:

[tex]K_f = K_i + W = 50 + (-10) = 40 J[/tex]