A roller coaster must do work raising its cars to the highest point on then ride. From there, the cars coast at

Question

varying speed until they return to the starting point. Suppose a loaded roller coaster car must be pulled
3.00 × 102 m from the ride’s starting point to the top of the first rise. If 2.13 × 106 J of work must be done on the
car during this stage of the ride, how large is the force exerted on the car by the raising mechanism?

Answer 1

To find the force exerted on the car by the raising mechanism, we can use the work formula:

\[ \text{Work} = \text{Force} \times \text{Distance} \cdot \cos(\theta) \]

In this case, the angle \(\theta\) is 0 degrees because the force applied by the raising mechanism is in the same direction as the movement of the car. Thus, \(\cos(0) = 1\), and the formula simplifies to:

\[ \text{Work} = \text{Force} \times \text{Distance} \]

Rearranging this formula to solve for Force gives us:

\[ \text{Force} = \frac{\text{Work}}{\text{Distance}} \]

Given that the work done \(W = 2.13 \times 10^6 , \text{J}\) and the distance \(d = 3.00 \times 10^2 , \text{m}\), we can now substitute these values into the equation:

\[ \text{Force} = \frac{2.13 \times 10^6 , \text{J}}{3.00 \times 10^2 , \text{m}} \]

Calculating this gives:

\[ \text{Force} = \frac{2.13 \times 10^6}{3.00 \times 10^2} = 7100 , \text{N} \]

Thus, the force exerted on the car by the raising mechanism is 7100 N.