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. square

Here's how to calculate the work done to slow down the truck:<br /><br />**1. Understand the concept:** 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 (energy is being removed from 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 />* 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:** -375,000 Joules. The negative sign indicates that work is done *on* the truck (to slow it down), rather than *by* the truck.<br />