Question

A force given by F(x) = 5x3 (in N/m3) acts on a 1-kg mass moving on a frictionless surface. The mass moves from x = 3.87 m to x = 6.09 m. a) How much work is done by the force? b) If the mass has a speed of 2 m/s at x = 3.87 m, what is its speed at x = 6.09 m?

Answer 1

To calculate the work done by a force described by F(x) = 5x^3 on a 1-kg mass moving on a frictionless surface and to determine the speed of the mass at a specific position, we can use the work-energy theorem and integrate the force over the displacement.

Step-by-step explanation:

To calculate the work done by a force, we need to integrate the force over the displacement. In this case, the force is given by F(x) = 5x^3 and the mass is 1 kg. The work done is given by the integral of F(x) with respect to x between the initial and final positions.

a) Work = ∫F(x) dx = ∫(5x^3) dx

b) To find the speed at x = 6.09 m, we can use the work-energy theorem. The work done by the force is equal to the change in kinetic energy of the mass.

Answer 2

The work done by the force F(x) = 5x^3 as a mass moves from x = 3.87 m to x = 6.09 m is calculated by integrating the force over the distance. Using the work-energy theorem, one can find the speed of the mass at x = 6.09 m given its initial velocity.

Step-by-step explanation:

The student asks how to calculate the work done by a force and the resulting speed of a mass after it has moved a certain distance on a frictionless surface. To find the work done, we integrate the force over the distance. Since F(x) = 5x3, the work done (W) from x = 3.87 m to x = 6.09 m is the integral of F(x) from 3.87 to 6.09, which gives W = (5/4)(x4)[from 3.87 to 6.09]. After finding the work done, we can use the work-energy theorem to determine the final speed of the mass at x = 6.09 m. The work-energy theorem states that the work done on the object is equal to the change in kinetic energy, so W = ½ m(vfinal2 - vinitial2).

Given the initial speed vinitial = 2 m/s at x = 3.87 m, the final speed vfinal at x = 6.09 m can be found by rearranging the work-energy theorem:

vfinal = √(vinitial2 + (2*W)/m)