Question

In Example 7.7, we found that the speed of a roller coaster that had descended 20.0 m was only slightly greater when it had an initial speed of 5.00 m/s than when it started from rest. This implies that Δ {PE} >> {KE}ᵢ ). Confirm this statement by taking the ratio of Δ {PE} to {KE}ᵢ ). (Note that mass cancels.)

A. frac{Δ {PE}}{{KE}ᵢ} = 10
B. frac{Δ {PE}}{{KE}ᵢ} = 20
C. frac{Δ {PE}}{{KE}ᵢ} = 30
D. frac{Δ {PE}}{{KE}ᵢ} = 40

Answer 1

The ratio of the change in gravitational potential energy (ΔPE) to the initial kinetic energy (KEi) of a roller coaster descending 20.0 meters confirms that ΔPE is significantly larger than KEi, verifying the statement that ΔPE >> KEi.

Step-by-step explanation:

The question is asking us to compare the change in gravitational potential energy (ΔPE) to the initial kinetic energy (KEi) of a roller coaster as it descends 20.0 meters. Given that the initial kinetic energy when the roller coaster has an initial speed of 5.00 m/s is calculated using the formula KEi = ½mv2 and the mass (m) cancels out, we have KEi = ½(5.00 m/s)2 = 12.5 J.

The change in potential energy (ΔPE) as the roller coaster descends h = 20.0 m can be calculated using the equation ΔPE = mgh, which simplifies to ΔPE = 9.81 m/s2 × 20.0 m = 196.2 J. The ratio of ΔPE to KEi is therefore ΔPE/KEi = 196.2 J / 12.5 J = 15.7, indicating that the change in gravitational potential energy is indeed much larger than the initial kinetic energy. This confirms that ΔPE >> KEi, consistent with the given statement.

Since mass cancels out, we can use the equation ΔPE = mgh for the change in potential energy and KE1 = 0.5mv2 for the initial kinetic energy. Substituting the given values, we have ΔPE = mgh = (mass)(gravity)(height) and KE1 = 0.5m(0)2 = 0. Therefore, the ratio of ΔPE to KE1 is ΔPE/KE1 = (mgh)/0 = undefined. This means that the change in potential energy is infinitely greater than the initial kinetic energy.