Find the work in joules needed to slow a 2500 kg truck from 20 m/s to 10m/s Hint: keep in mind the truck is slowing down.

Here's how to calculate the work done to slow down the truck:<br /><br />**1. Understand the concept:**<br /><br />Work done on an object is equal to the change in its kinetic energy. Since the truck is slowing down, the work done is negative (meaning work is done *by* the truck, not *on* the truck).<br /><br />**2. Formula:**<br /><br />Work (W) = Change in Kinetic Energy (ΔKE)<br /><br />ΔKE = KE_final - KE_initial<br /><br />KE = (1/2) * m * v^2<br /><br />where:<br /><br />* m = mass (kg)<br />* v = velocity (m/s)<br /><br />**3. Calculate the initial kinetic energy:**<br /><br />KE_initial = (1/2) * 2500 kg * (20 m/s)^2 = 500,000 J<br /><br />**4. Calculate the final kinetic energy:**<br /><br />KE_final = (1/2) * 2500 kg * (10 m/s)^2 = 125,000 J<br /><br />**5. Calculate the change in kinetic energy (and therefore the work):**<br /><br />W = ΔKE = KE_final - KE_initial = 125,000 J - 500,000 J = -375,000 J<br /><br />**Answer:**<br /><br />The work done to slow the truck is -375,000 Joules. The negative sign indicates that the truck is doing work on its surroundings (e.g., through friction with the brakes and road) to slow down. 375,000 Joules of kinetic energy are converted to other forms of energy (primarily heat).<br />