A 7.0-kg rock is subject to a variable force given by the equation f(x)=6.0n−(2.0n/m)x+(6.0n/m2)x2 if the rock initially is at rest at the origin, find its speed when it has moved 9.0 m.

For Newton's second law, the force is equal to the product between the mass and the acceleration of the rocket:
[tex]F=ma[/tex]
From which we can rewrite the acceleration as
[tex]a(x)= \frac{F}{m} = \frac{1}{m} (6-2x+6x^2)[/tex]
where m=7.0 kg.

The velocity of the rocket is the derivative of the acceleration:
[tex]v(x) = \frac{1}{m} (-2+12 x) [/tex]
and if we substitute x=9.0 m, we find the rocket velocity after 9.0 m:
[tex]v(9)= \frac{1}{7}(-2+12\cdot 9)=15.1 m/s [/tex]

batolisis batolisis · Answer 2 · 2022-03-22T13:18:36+01:00

The speed of the rock after it has moved 9.0 m = 20.22 m/s

Given data:

mass of rock ( m ) = 7 kg

variable force : 6.0 N - ( 2.0 N/m )x + (6.0N/m²) x²

Determine the speed of the rock

To determine the speed of the rock after it has moved 9 m we will integrate the variable force

F(x) = 6 - 2x + 6x²

therefore :

[tex]m\int\limits^v_0 {vdv} \, = \int\limits^9_0 {(6x^2-2x+6)} \, dx[/tex]

7 * [tex]\frac{v^2}{2}[/tex] = [ 2x³ - x² + 6x ]₀⁹

7v²/2 = 1458 - 81 + 54

v = 20.22 m/s

Hence we can conclude that The speed of the rock after it has moved 9.0 m = 20.22 m/s

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