If -0.750J of work is done on the book from B to C, how fast is it moving at point C?

Steps to Solve

1. Identify initial conditions The initial speed of the book at point B is given as $v_B = 1.25 , \text{m/s}$.

2. Use the work-energy principle According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. The equation for kinetic energy is given by: $$ KE = \frac{1}{2} mv^2 $$

3. Calculate initial kinetic energy at point B Using the mass of the book ($m = 1.50 , \text{kg}$) and its speed at point B, we calculate: $$ KE_B = \frac{1}{2} \cdot 1.50 , \text{kg} \cdot (1.25 , \text{m/s})^2 $$

4. Compute the change in kinetic energy due to work done The work done on the book from B to C is $W = -0.750 , \text{J}$. Therefore, the final kinetic energy at point C is: $$ KE_C = KE_B + W $$

5. Set up the equation for final kinetic energy Express the final kinetic energy at point C in terms of its speed $v_C$: $$ KE_C = \frac{1}{2} mv_C^2 $$

6. Equate the two expressions for kinetic energy Setting the two expressions for $KE_C$ equal gives us: $$ \frac{1}{2} mv_C^2 = KE_B - 0.750 , \text{J} $$

7. Solve for the speed at point C Plugging in the values and rearranging the equation will yield $v_C$: $$ v_C = \sqrt{\frac{2 \cdot (KE_B - 0.750 , \text{J})}{m}} $$