Solving for v, we getv=\sqrt{\frac{2}{m}\int_{x_1} ^{x_2} F dx}

[ssm11s]

1. A 2 kg book is at rest on a flat frictionless surface at initial position xi=0. A force F= (2.5-x²) is applied on the book. What is the Kinetic Energy of the block as it passes through x=2 m ?



2. W= Int ( force ) ; K= (1/2)mv² ; W = K₁ - K₂



3. So I started out with finding the W by taking the integral of F from x=0 to x=2. Then i put the value of the work in the work-kinetic energy equation. And I am stuck here cause i don't know where to use the mass of the book.

 

[rock.freak667]

Well now, the value of your integral is the kinetic energy.

$$Change \ in \ kinetic \ energy = \int_{x_1} ^{x_2} F dx$$


Since it started at rest, the initial ke is 0 and the final one is 1/2mv²

so

$$\frac{1}{2}mv^2-0=\int_{x_1} ^{x_2} F dx$$