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Chapter 7: Problem 1

What average force is necessary to bring a 50.0-kg sled from rest to a speed of \(3.0 \mathrm{m} / \mathrm{s}\) in a period of \(20.0 \mathrm{s} ?\) Assume frictionless ice.

Short Answer

Expert verified
Answer: The average force necessary is 7.5 N.

Step by step solution

01

Identify the given values

The problem provides us with the following values: - Mass (m) = 50.0 kg - Initial velocity (u) = 0 m/s (sled is initially at rest) - Final velocity (v) = 3.0 m/s - Time (t) = 20.0 s
02

Calculate the acceleration

To calculate the acceleration (a), we will use the formula: \(a = \frac{v-u}{t}\) Plugging in the given values, we get: \(a = \frac{3.0-0}{20.0}\) \(a = \frac{3.0}{20.0}\) \(a = 0.15 \, \mathrm{m/s^2}\)
03

Calculate the average force

Now that we have the acceleration value, we can use Newton's second law (F = ma) to find the average force (F): \(F = m \times a\) \(F = 50.0 \, \mathrm{kg} \times 0.15 \, \mathrm{m/s^2}\) \(F = 7.5\, \mathrm{N}\) So, the average force necessary to bring the 50.0-kg sled from rest to a speed of \(3.0 \, \mathrm{m/s}\) in a period of \(20.0\, \mathrm{s}\) on frictionless ice is \(7.5\, \mathrm{N}\).

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