Find Work of a Force Acted on Mass over Frictionless Surface

[mrsteve]

Homework Statement

A force given by $$F(x)=5x^3 (in \frac {N}{m^3})$$ acts on a 1-kg mass moving on a frictionless surface. The mass moves from x = 2m to x = 6m.

a) How much work is done by the force?

Homework Equations

$$W=\int F_x from x_o to x_f$$

The Attempt at a Solution

$$\int 5x^3 from x=2 to x=6 = 1.6*10^3$$

Is this correct?

 

[Ignea_unda]

Looks correct to me, mrsteve.

 

[mrsteve]

Also, if the speed of the mass is 2 m/s @ x=2 what is the speed at x=6?

$$KE=\frac{mv^2}{2}$$

$$1600=\frac{(1)(v^2)}{2} => v=5.6*10^1 = 60 m/s$$

Would that be correct?

 

[mrsteve]

Looks correct to me, mrsteve.

The book has 2,000 J. I suppose that's because there's only 1 significant digit in mass = 1kg?

$$1.6 *10^3 => 2 *10^3$$

 

[rwisz]

Yes, your solution is correct. The work done by a force is given by the integral of the force over the displacement, which in this case is from x=2 to x=6. Since the force is given by F(x)=5x^3, the integral becomes \int 5x^3 from x=2 to x=6, which equals 1.6*10^3. This means that the force has done 1.6*10^3 Joules of work on the mass.