A force given by F(x) = 4x^3 x̂ (in N/m^3) acts on a 1kg mass moving on a frictionless surface. The mass moves from x = 4 m to x = 7 m.
(a) How much work is done by the force?
(b) If the mass has a speed of 1 m/s at x = 4 m, what is its speed at x = 7 m?
work done = integral of F dx
= 4* integral from 4 to 7 of x^3 dx
= 4 * (7^4/4 - 4^4/4)
= 7^4 - 4^4
= 2145 Joules
=================
(1/2) m v^2 = 2145 + (1/2)m(1^2)
or in other words
final kinetic energy = work in + initial kinetic energy.
(a) To calculate the work done by the force, we first need to determine the equation for work done. The work done by a force is given by the formula:
W = ∫ F(x) · dx
where F(x) is the force as a function of position and dx represents the displacement.
In this case, F(x) = 4x^3 x̂ and the mass moves from x = 4 m to x = 7 m.
We can find the work done by integrating the force with respect to x over the given range:
W = ∫[4x^3 x̂] · dx
= ∫[4x^3] · dx
= ∫4x^3 · dx
= 4 ∫x^3 · dx
Using the power rule for integration, we can integrate x^3 as follows:
W = 4 * (x^4/4) + C
= x^4 + C
Evaluated over the given range from x = 4 m to x = 7 m:
W = (7^4 + C) - (4^4 + C)
= 2401 - 256
= 2145 J
Therefore, the work done by the force is 2145 joules (J).
(b) To determine the speed of the mass at x=7m, we can use the work-energy principle. The work done by a force is equal to the change in kinetic energy of the mass.
Therefore, we can write:
W = ΔKE
Since the force is conservative (independent of path), we can assume that the work done is solely due to a change in the gravitational potential energy (PE).
W = ΔPE
Using the equation for potential energy, PE = mgh (where h is the height above a reference point), and rearranging the equation, we find:
W = mgh
The work done is also equal to the change in kinetic energy (ΔKE):
W = ΔKE
Since the mass is moving on a frictionless surface, there is no work done by the force of friction. Therefore, the work done by the force is equal to the change in kinetic energy.
W = ΔKE
Since work is equal to the change in kinetic energy, we can write:
KE_final - KE_initial = W
Using the given information, the mass is 1 kg, and its initial speed is 1 m/s at x = 4 m. Therefore:
KE_initial = (1/2)mv^2 = (1/2)(1 kg)(1 m/s)^2 = 0.5 J
The work done by the force is 2145 J, as calculated in part (a).
So, using the equation above, we can determine the final kinetic energy:
KE_final - 0.5 J = 2145 J
Substituting the values:
KE_final - 0.5 J = 2145 J
KE_final = 2145 J + 0.5 J
= 2145.5 J
The final kinetic energy is 2145.5 J.
To find the final speed, we can use the equation for kinetic energy:
KE = (1/2)mv^2
Rearranging the equation, we find:
v^2 = 2KE/m
Substituting the given values:
v^2 = (2)(2145.5 J)/(1 kg)
v^2 = 4291 J/kg
v = √4291 J/kg ≈ 65.52 m/s
Therefore, the speed of the mass at x = 7 m is approximately 65.52 m/s.
To find the work done by the force, you can use the formula:
Work = ∫F(x) · dx
where F(x) is the force and dx is the infinitesimal change in position.
(a) Work done by the force:
Given F(x) = 4x^3 x̂ and the mass is moving from x = 4 m to x = 7 m.
We can integrate F(x) · dx from x = 4 m to x = 7 m:
Work = ∫F(x) · dx from 4 to 7
= ∫(4x^3) · dx from 4 to 7
= ∫(4x^3) dx from 4 to 7
= [x^4] from 4 to 7
= (7^4) - (4^4)
= 2401 - 256
= 2145 J
Therefore, the work done by the force is 2145 Joules.
(b) To find the speed at x = 7 m, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy:
Work = ΔKE
Since there is no change in potential energy and no work done by any other forces, the work done by the force F(x) is equal to the change in kinetic energy of the mass.
So, we can set up the equation:
Work = ΔKE
2145 J = (1/2)mv^2 - (1/2)m(1 m/s)^2
Here, m = 1 kg (mass of the object) and v is the speed at x = 7 m.
Rearranging the equation, we get:
(1/2)v^2 = 2145 J + (1/2)(1 kg)(1 m/s)^2
(1/2)v^2 = 2145 J + 0.5 J
(1/2)v^2 = 2145.5 J
v^2 = 4291 J
v = √4291 J
v ≈ 65.57 m/s
Therefore, the speed at x = 7 m is approximately 65.57 m/s.