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

Steps to Solve

1. Identify Given Values

We have the following information:

• Mass of the book, $m = 1.50 , \text{kg}$
• Initial speed at point B, $v_B = 1.25 , \text{m/s}$
• Work done on the book, $W = -0.750 , \text{J}$

2. Calculate Initial Kinetic Energy at Point B

The kinetic energy (KE) at point B can be calculated using the formula: $$ KE_B = \frac{1}{2} m v_B^2 $$

Substituting the values: $$ KE_B = \frac{1}{2} \times 1.50 , \text{kg} \times (1.25 , \text{m/s})^2 $$

3. Calculate the Kinetic Energy at Point C

According to the work-energy principle, the kinetic energy at point C (KE_C) can be found from the kinetic energy at point B and the work done on the book: $$ KE_C = KE_B + W $$

4. Calculate Final Speed at Point C

We can find the final speed at point C using the relationship between kinetic energy and speed: $$ KE_C = \frac{1}{2} m v_C^2 $$

Rearranging for $v_C$, we have: $$ v_C = \sqrt{\frac{2 KE_C}{m}} $$

5. Substitute and Calculate Final Speed

Substituting the values we calculated for $KE_C$ and $m$ to find $v_C$.