Question

In: Physics

The only force acting on a 8 kg body as it moves along the x axis...

The only force acting on a 8 kg body as it moves along the x axis varies as shown in the figure. The velocity of the body at x = 0 is 8.0 m/s. What is the kinetic energy of the body at x =3.0 m?

At what value of x will the body have a kinetic energy of 244.40 J?

What is the maximum kinetic energy attained by the body between x = 0 and x = 5.0 m

Solutions

Expert Solution

Let:
F be the variable force,
m be the mass of the body,
v be its initial velocity,
K be the kinetic energy at displacement x,
W be the work done by F.

From the graph:
F = 4 – 4X N. until x = 2, and then F = - 4N.

For x <= 2m:
K = mv2 / 2 + int(0, x) [ 4 - 4x ] dx
= mv2 / 2 + 4x - 2x2
K= 256 + 4x - 2x2 ----(1)

At x = 2.0m, K = mv2 / 2.

For x > 2.0m,
K = mv2 / 2 - int(2, x) [ 4 ] dx
= mv2 / 2 - (4x - 8)
= mv2 / 2 - 4x + 8
K= 256 - 4x + 8 ----(2)


a) the kinetic energy of body at x=3.0m
At x = 3.0m:
K = 256 - 4 x 3 + 8
= 252 J.


b) the value of x will the body have a kinetic energy of 244.40 J?

Putting K = 244.40 in (1)
244.4 = 256 + 4x - 2x2
2x2 - 4x - 11.6 = 0
x = [ 4 +/- √(42 +( 4 x 2 x 11.6) ] / (2 x 2)
= 3.6m or - 1.6m.

Putting K = 244.4 in (2)
244.4 = 256 - 4x + 8
x = 4.9 m.

(c)
When K is maximum at a stationary point, its derivative is 0.


256 + 4 - 4x for x <= 2.0m ----(3)
256 - 4 for x > 2.0m ---(4)

dK / dx = 0 when:
256 + 4 - 4x = 0
x = 65 m.
the maximum KE in [0, 5.0m] is at x = 2.0m, and its value is 256 J.


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