A roller coaster is moving at \(2.00 \mathrm{~m} / \mathrm{s}\) at the top of the first hill \((h=40.0 \mathrm{~m}) .\) Ignoring friction and air resistance, how fast will the roller coaster be moving at the top of a subsequent hill, which is \(15.0 \mathrm{~m}\) high?

First, we need to find the initial kinetic energy (KE) and potential energy (PE) of the roller coaster at the top of the first hill. The initial kinetic energy can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass of the roller coaster and v is its speed. We are not given the mass of the roller coaster, but since the mass remains the same throughout the problem, we can work with the expression 0.5 * m * v^2. The initial potential energy can be calculated using the formula PE = m * g * h, where g is the acceleration due to gravity (9.81 m/s²) and h is the height of the first hill. So we have: Initial KE = 0.5 * m * (2.00 m/s)^2 Initial PE = m * (9.81 m/s²) * (40.0 m)

According to the conservation of mechanical energy principle, the sum of the kinetic and potential energies of the roller coaster at the top of the first and second hills should be equal. So we can write: Initial KE + Initial PE = Final KE + Final PE At the top of the second hill, the height is 15.0 m, so we can calculate the final potential energy: Final PE = m * (9.81 m/s²) * (15.0 m) Now, we can substitute the expressions for initial and final potential energy into the conservation of mechanical energy equation: 0.5 * m * (2.00 m/s)^2 + m * (9.81 m/s²) * (40.0 m)= Final KE + m * (9.81 m/s²) * (15.0 m) We can solve this equation for the final kinetic energy: Final KE = 0.5 * m * (2.00 m/s)^2 + m * (9.81 m/s²) * (40.0 m) - m * (9.81 m/s²) * (15.0 m)

Now that we have the expression for the final kinetic energy, we can calculate the final speed of the roller coaster. The final KE can be written as: Final KE = 0.5 * m * v_f^2 where v_f is the final speed we want to find. Solving this equation for v_f, we have: v_f^2 = 2 * (0.5 * m * (2.00 m/s)^2 + m * (9.81 m/s²) * (40.0 m) - m * (9.81 m/s²) * (15.0 m))/(m) We can see that m, the mass of the roller coaster, cancels out: v_f^2 = 2 * (0.5 * (2.00 m/s)^2 + (9.81 m/s²) * (40.0 m) - (9.81 m/s²) * (15.0 m)) Now, we can calculate the final speed: v_f = sqrt(2 * (0.5 * (2.00 m/s)^2 + (9.81 m/s²) * (40.0 m) - (9.81 m/s²) * (15.0 m))) v_f ≈ 23.96 m/s Thus, the roller coaster will be moving at approximately 23.96 m/s at the top of the second hill, which is 15.0 m high.