Translational Kinetic Energy — Practice Problem Solutions

1. $K=\tfrac12mv^2=\tfrac12(2.0)(3.0)^2=\tfrac12(2.0)(9.0)=9.0\text{ J}$.

2. Zero. At the peak, the ball's velocity is exactly zero, and $K=\tfrac12m(0)^2=0$ — kinetic energy can never be negative, and it's exactly zero here, not negative, even though gravity is still acting on the ball at that same instant.

3. $K_f=\tfrac12(0.20)(5.0)^2=2.5\text{ J}$. Since the object started from rest, $K_i=0$, so $\Delta K=2.5\text{ J}-0=2.5\text{ J}$.

4. $v=\dfrac{\Delta x}{\Delta t}=\dfrac{(0-8)}{(5-4)}=-8\text{ m/s}$; speed is $8\text{ m/s}$. $K=\tfrac12(5)(8)^2=\tfrac12(5)(64)=160\text{ J}$.

5. Decreasing. The projectile's speed is decreasing on the way up (gravity's acceleration opposes its motion), and since $K\propto v^2$, a decreasing speed always means decreasing kinetic energy.

6. a) $K=\tfrac12(1300)(25)^2=406{,}250\text{ J}\approx4.1\times10^5\text{ J}$. b) Relative speed $=25-15=10\text{ m/s}$; $K=\tfrac12(1300)(10)^2=65{,}000\text{ J}=6.5\times10^4\text{ J}$. c) Kinetic energy depends on speed, and speed is relative to the observer — the car's speed relative to the ground (25 m/s) differs from its speed relative to an observer in the truck (10 m/s), because the truck itself is moving, so the two kinetic energy values are both correct for their respective observers.

7. a) $K_i=0$ at $t=0$; $K_f=\tfrac12(0.20)(5)^2=2.5\text{ J}$ (reached at $t=2\text{ s}$ and held). $\Delta K=2.5\text{ J}-0=2.5\text{ J}$. b) No — velocity is constant at 5 m/s during the last 3 seconds, so kinetic energy is also constant (at 2.5 J) during that interval.

8. (A) is correct. Kinetic energy depends only on mass and speed, $K=\tfrac12mv^2$ — direction plays no role. Since the mass and speed are the same at $t_1$ and $t_2$ (only the direction changed, west to east), $\Delta K=0$.