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A girl and her bicycle have a total mass of 40 kg . At the top of the hill her speed is \( 5.0 \mathrm{~m} / \mathrm{s} \). The hill is 10 m high and 100 m long. If the force of friction as she rides down the hill is 20 N , what is her speed at the bottom? 1) \( 5.0 \mathrm{~m} / \mathrm{s}^{\text {- }} \) 2) \( 10 \mathrm{~m} / \mathrm{s} \) 3) \( 11 \mathrm{~m} / \mathrm{s} \) 4) \( 14 \mathrm{~m} / \mathrm{s} \) 5) She stops before she reaches the bottom.

Ask by Hart May. in the United States
Feb 25,2025

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Her speed at the bottom is 11 m/s.

Solución

To solve the problem, we use energy conservation accounting for friction. The total energy input available is the initial kinetic energy plus the gravitational potential energy lost by descending 10 m. Some energy is "lost" due to the work done against friction. 1. Calculate the initial kinetic energy:  KE_initial = (1/2) m v₀² = 0.5 × 40 kg × (5.0 m/s)² = 0.5 × 40 × 25 = 500 J 2. Calculate the gravitational potential energy lost:  ΔPE = m g h = 40 kg × 9.8 m/s² × 10 m = 3920 J 3. Determine the work done against friction:  W_friction = friction force × distance = 20 N × 100 m = 2000 J 4. Total mechanical energy at the bottom (assuming no other energy losses) is:  E_bottom = KE_initial + ΔPE − W_friction  E_bottom = 500 J + 3920 J − 2000 J = 2420 J 5. At the bottom, all of this energy is kinetic:  (1/2) m v² = 2420 J ⟹ (1/2) × 40 kg × v² = 2420 J  20 v² = 2420 ⟹ v² = 2420 / 20 = 121  v = √121 = 11 m/s Thus, her speed at the bottom of the hill is 11 m/s, which corresponds to option 3.

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To find the girl's speed at the bottom of the hill, we can use the principle of conservation of energy along with the work done against friction. The initial gravitational potential energy is converted into kinetic energy and work done against friction, which can be calculated as follows: 1. The initial potential energy at the top is \( PE = mgh = 40 \times 9.81 \times 10 \approx 3924 \, \text{J} \). 2. The kinetic energy at the top is \( KE = \frac{1}{2} mv^2 = \frac{1}{2} \times 40 \times (5)^2 = 500 \, \text{J} \). 3. The work done against friction is \( W = F \times d = 20 \times 100 = 2000 \, \text{J} \). The total mechanical energy at the top minus the work done against friction gives us the kinetic energy at the bottom: \[ KE_{bottom} = (PE + KE_{top}) - W = (3924 + 500 - 2000) = 2424 \, \text{J}. \] Now, we can calculate the speed at the bottom using the kinetic energy formula: \[ KE_{bottom} = \frac{1}{2} mv^2 \Rightarrow 2424 = \frac{1}{2} \times 40 \times v^2 \Rightarrow v^2 = \frac{2424 \times 2}{40} = 121.2 \Rightarrow v \approx 11.0 \, \text{m/s}. \] So the answer is: 3) \( 11 \mathrm{~m} / \mathrm{s} \) Let's add some extra insight! First, riding down a hill not only gives you a speed boost thanks to gravity but also can feel like flying, especially when the wind whips through your hair! Just remember, it’s not just about speed; managing your bike’s balance and grip on the road is crucial too. Also, did you know many cyclists practice braking efficiently? It’s a mix of using both front and rear brakes appropriately to maintain control while descending steep hills, and to avoid flipping over the handlebars. So next time you ride, think about mastering those brakes – it can make your rides much more enjoyable!

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