Worked example 5.4: Roller coaster ride

Question: A roller coaster cart of mass $m= 300 {\rm kg}$ starts at rest at point $A$, whose height off the ground is $h_1=25 {\rm m}$, and a little while later reaches point $B$, whose height off the ground is $h_2=7 {\rm m}$. What is the potential energy of the cart relative to the ground at point $A$? What is the speed of the cart at point $B$, neglecting the effect of friction?

Answer: The gravitational potential energy of the cart with respect to the ground at point $A$ is

$$U_A = m g h_1 = 300\times 9.81 \times 25 = 7.36\times 10^4 {\rm J}.$$

Likewise, the potential energy of the cart at point $B$ is

$$U_B = m g h_2 = 300\times 9.81 \times 7 = 2.06\times 10^4 {\rm J}.$$

Hence, the change in the cart's potential energy in moving from point $A$ to point $B$ is

$${\mit\Delta} U = U_B-U_A = 2.06\times 10^4-7.36\times 10^4 = -5.30\times 10^4 {\rm J}.$$

By energy conservation, ${\mit\Delta}K = -{\mit\Delta} U$, where $K$ represents kinetic energy. However, since the initial kinetic energy is zero, the change in kinetic energy ${\mit\Delta} K$ is equivalent to the final kinetic energy $K_B$. Thus,

$$K_B = -{\mit\Delta}U = 5.30\times 10^4 {\rm J}.$$

Now, $K_B = (1/2) m v_B^{ 2}$, where $v_B$ is the final speed. Hence,

$$v_B = \sqrt{\frac{2 K_B}{m}} = \sqrt{\frac{2\times 5.30\times 10^4}{300}} = 18.8 {\rm m/s}.$$