Question

A roller coaster is towed up an incline at a steady speed of $0.50 \mathrm{m} / \mathrm{s}$ by a chain parallel to the surface of the incline. The slope is \(3.0 \%,\) which means that the elevation increases by \(3.0 \mathrm{m}\) for every \(100.0 \mathrm{m}\) of horizontal distance. The mass of the roller coaster is \(400.0 \mathrm{kg}\). Ignoring friction, what is the magnitude of the force exerted on the roller coaster by the chain?

Short answer

Answer: The force exerted by the chain is 117.72 N.

Step-by-step solution

Calculate the angle of incline

Since the slope is given as \(3.0 \%\), we can find the tangent of the angle of incline, which is the ratio of elevation to horizontal distance. \(\tan(\theta) = \frac{3.0}{100.0}\). Now we can find the angle of incline using the arctangent function: \(\theta = \arctan\left(\frac{3}{100}\right)\).

Calculate the gravitational force acting on the roller coaster

The gravitational force acting on the roller coaster, \(F_g\), can be calculated using \(F_g = mg\), where \(m\) is the mass of the roller coaster, and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m/s^2}\)). So, \(F_g = 400.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 3924 \mathrm{N}\).

Calculate the component of gravitational force parallel to the incline

Since the force exerted by the chain opposes the component of gravitational force parallel to the incline, which we can find as \(F_{g\parallel} = F_g \sin(\theta)\), using the angle we found in Step 1. We find: \(F_{g\parallel} = 3924 \mathrm{N} \times \sin(\arctan(0.03)) = 117.72 \mathrm{N}\).

Apply Newton's second law

The force exerted by the chain, \(F_c\), must equal the component of gravitational force parallel to the incline since the roller coaster is moving at a constant speed (no acceleration). Using Newton's second law \(\Sigma \boldsymbol{F} = m \boldsymbol{a}\), we can write: \(F_c - F_{g\parallel} = ma \implies F_c = F_{g\parallel}\), since the acceleration is 0. \(F_c = 117.72 \mathrm{N}\) Therefore, the magnitude of the force exerted on the roller coaster by the chain is \(117.72 \mathrm{N}\).