A 1.40 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as F = 3x^2 - 2x. If the block's speed passing through the origin was 3.80 m/s, with what speed (in meters/second) does it pass the point x = 8.00 m?
To find the speed of the block at a given point, we need to use the principle of conservation of energy. Since the surface is frictionless, the only force acting on the block is the varying force F = 3x^2 - 2x.
The potential energy associated with this force can be defined as the negative of the integral of the force over the distance traveled. The potential energy is given by the equation:
PE = - ∫ F dx
Integrating F = 3x^2 - 2x with respect to x, we get:
PE = - ∫ (3x^2 - 2x) dx
To find the potential energy at point x = 8.00 m, we evaluate the integral:
PE = - ∫ (3x^2 - 2x) dx from 0 to 8
PE = - [x^3 - x^2] evaluated from 0 to 8
PE = - [(8^3 - 8^2) - (0^3 - 0^2)]
PE = - [512 - 64]
PE = - 448 Joules
Now, using the principle of conservation of energy, we know that the total mechanical energy of the block remains constant. It is given by the sum of its kinetic energy (KE) and potential energy (PE):
Total Mechanical Energy (E) = KE + PE
Since there is no potential energy at the point of origin, the total mechanical energy at that point is equal to the kinetic energy:
E(origin) = KE(origin)
Plugging in the speed at the point of origin (x = 0) and the mass of the block, we get:
0.5 * m * v^2 = 0.5 * 1.40 kg * (3.80 m/s)^2
0.5 * m * v^2 = 3.192 J
Now, we can find the kinetic energy (KE) at the point x = 8.00 m using the same equation:
E(x = 8) = KE(x = 8) + PE(x = 8)
0.5 * m * v^2 = - 448 J + KE(x = 8)
Plugging in the calculated potential energy and rearranging the equation, we get:
KE(x = 8) = 0.5 * 1.40 kg * (3.192 J + 448 J)
KE(x = 8) = 1.9856 J
Finally, to find the speed v at the point x = 8.00 m, we use the equation:
0.5 * m * v^2 = KE(x = 8)
0.5 * 1.40 kg * v^2 = 1.9856 J
Dividing both sides by 0.7 kg, we get:
v^2 = 2.8394 m^2/s^2
Taking the square root, we find:
v = √(2.8394 m^2/s^2)
v ≈ 1.68 m/s
Therefore, the block passes the point x = 8.00 m with a speed of approximately 1.68 m/s.