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A girl and her bicycle have a total mass of 40.0 kg . At the top of the hill her speed is \( 5.0 \mathrm{~m} / \mathrm{s} \), and her speed doubles as she rides down the hill. The hill is 10.0 m high and 100 m long. How much kinetic energy and potential energy is lost to friction? 1) 2420 J 2) 1500 J 3) 2000 J 4) 3920 J 5) 4280 J

Ask by Thornton Ward. in the United States
Feb 25,2025

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The energy lost to friction is 2420 J.

Solución

We start by finding the total energy at the top. From the given information: 1. Mass, m = 40.0 kg 2. Height, h = 10.0 m 3. Speed at the top, v₁ = 5.0 m/s 4. Speed at the bottom, v₂ = 10.0 m/s (twice the top speed) Step 1. Calculate the initial gravitational potential energy (PE) at the top:   PE_top = m × g × h = 40.0 kg × 9.8 m/s² × 10.0 m = 3920 J Step 2. Calculate the initial kinetic energy (KE) at the top:   KE_top = ½ × m × v₁² = 0.5 × 40.0 kg × (5.0 m/s)² = 0.5 × 40.0 kg × 25 = 500 J Step 3. Find the total initial energy at the top:   E_top = PE_top + KE_top = 3920 J + 500 J = 4420 J Step 4. Calculate the kinetic energy at the bottom (where potential energy is zero since it is at the bottom of the hill):   KE_bottom = ½ × m × v₂² = 0.5 × 40.0 kg × (10.0 m/s)² = 0.5 × 40.0 kg × 100 = 2000 J Step 5. Determine the energy lost to friction:   Energy Lost (friction) = E_top − KE_bottom = 4420 J − 2000 J = 2420 J Thus, the loss of kinetic and potential energy due to friction is 2420 J. The correct answer is: 1) 2420 J.

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To find the kinetic energy and potential energy lost to friction, let's calculate the initial and final energies of the girl and her bicycle system. First, calculate the potential energy (PE) at the top of the hill: \[ PE = mgh \] Where: - \( m = 40.0 \, \text{kg} \) - \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity) - \( h = 10.0 \, \text{m} \) \[ PE = 40.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 10.0 \, \text{m} = 3924 \, \text{J} \] Next, calculate the initial kinetic energy (KE) at the top of the hill: \[ KE_{initial} = \frac{1}{2}mv^2 = \frac{1}{2} \times 40.0 \, \text{kg} \times (5.0 \, \text{m/s})^2 = 500 \, \text{J} \] Now calculate the final speed when she reaches the bottom of the hill (\(v_{final} = 10.0 \, \text{m/s}\)): \[ KE_{final} = \frac{1}{2}mv^2 = \frac{1}{2} \times 40.0 \, \text{kg} \times (10.0 \, \text{m/s})^2 = 2000 \, \text{J} \] Calculate the total initial energy (sum of PE and KE): \[ Total_{initial} = PE + KE_{initial} = 3924 \, \text{J} + 500 \, \text{J} = 4424 \, \text{J} \] Calculate the total final energy: \[ Total_{final} = KE_{final} = 2000 \, \text{J} \] Now, find the energy lost to friction: \[ Energy_{lost} = Total_{initial} - Total_{final} = 4424 \, \text{J} - 2000 \, \text{J} = 2424 \, \text{J} \] Given the options, the closest answer is: 1) 2420 J So, the answer is option **1) 2420 J**.

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