Question
Name: __ 1. A 1260 g apple is dropped from the top of a 2700m high tower. If there is a frictional force of 4 Nacting up on it, determine how long it will take to reach a speed of 20m/s
Solution
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Ferdinand
Master · Tutor for 5 years
Answer
Name: ______1. **Understanding the Problem:** An apple is falling from a great height, experiencing both gravity and air resistance (friction). We need to find the time it takes to reach a specific speed.2. **Known Values:*** Mass (m) = 1260 g = 1.26 kg* Initial velocity (v₀) = 0 m/s (since it's dropped)* Final velocity (v) = 20 m/s* Height of tower = 2700 m (This information is extraneous for this particular problem since we are only concerned with the time it takes to reach 20 m/s, not the ground.)* Frictional force (F_f) = 4 N* Acceleration due to gravity (g) = 9.8 m/s²3. **Finding the Net Force:**The apple experiences two forces: gravity pulling it down and friction acting upwards. The net force (F_net) is the difference between these two:F_net = Force due to gravity - Frictional forceF_net = (m * g) - F_fF_net = (1.26 kg * 9.8 m/s²) - 4 NF_net = 12.348 N - 4 NF_net = 8.348 N4. **Calculating the Acceleration:**Newton's second law of motion states F_net = m * a, where 'a' is the acceleration. We can rearrange this to solve for 'a':a = F_net / ma = 8.348 N / 1.26 kga ≈ 6.625 m/s²5. **Finding the Time:**We can use the following equation of motion to find the time (t) it takes to reach the final velocity:v = v₀ + a*tSince the initial velocity is 0, the equation simplifies to:v = a*tRearranging to solve for time:t = v / at = 20 m/s / 6.625 m/s²t ≈ 3.02 s**Answer:** It will take approximately 3.02 seconds for the apple to reach a speed of 20 m/s.