A particle of mass 150 grams was projected at 13 metres per second across a horizontal plane. It decelerated uniformly at two metres per second squared. Find the change in its kinetic energy in the first four seconds of motion.
We’re told in this statement that the particle has a mass of 150 grams. We’ll call that mass 𝑚. We’re also told that it has an initial speed of 13 metres per second. We’ll call the 𝑣 sub 𝑖. The particle undergoes a deceleration of two metres per second squared. We’ll call that value 𝑎. We want to find the change in the kinetic energy of the particle over the first four seconds of motion. We can call this change ΔKE.
To begin our solution, let’s recall that the kinetic energy of an object is equal to one-half the mass of that object times its speed squared. Since the change in kinetic energy of our particle ΔKE is equal to its final kinetic energy minus its initial kinetic energy, the equation for kinetic energy tells us we can write this as one-half 𝑚𝑣𝑓 squared, where 𝑣𝑓 is the particle’s final speed, minus one-half 𝑚𝑣𝑖 squared or factoring out in 𝑚 over two one-half 𝑚 times 𝑣𝑓 squared minus 𝑣𝑖 squared.
We know the particle’s mass 𝑚 as well as its initial speed 𝑣 sub 𝑖. But we want to solve for 𝑣 sub 𝑓. To solve for 𝑣 sub 𝑓, we can use the fact that acceleration is equal to the change in speed over the change in time. In our case, we can write that 𝑎 is equal to 𝑣 sub 𝑖 minus 𝑣 sub 𝑓, the final speed of the particle, divided by the time over which the particle decelerates, which is given in the problem statement as four seconds.
Rearranging this equation to solve for 𝑣 sub 𝑓, we see that 𝑣 sub 𝑓 is equal to 𝑣 sub 𝑖 minus 𝑎 times 𝑡 or plugging in for 𝑣 sub 𝑖, 𝑎, and 𝑡, we calculate 𝑣 sub 𝑓 to be five metres per second. That’s the speed of the particle after four seconds have passed.
We can now return to our ΔKE equation and we’re now prepared to plug in for each of the variables in this expression. When we do, being careful to write our mass in units of kilograms to be consistent with the units in the rest of our expression, we enter these values on our calculator and find that ΔKE is equal to negative 10.8 joules. That’s the change in kinetic energy this particle experiences.