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

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To find the kinetic energy of a 2 kg book on a frictionless surface when a force F = (2.5 - x^2) is applied, the work done by the force must be calculated using the integral from x=0 to x=2. The work-energy principle states that the work done is equal to the change in kinetic energy. Since the book starts at rest, the initial kinetic energy is zero, leading to the equation (1/2)mv^2 = ∫(from 0 to 2) F dx. The integral result directly provides the kinetic energy at x=2 m. This approach effectively utilizes the mass of the book in the final kinetic energy calculation.
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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.
 
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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
 
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