Suppose a horizontal force of 700 N is required to push a 200 kg crate across a warehouse floor at a constant speed of 2.5 m/s. (a) What is the power of this force? (b) If the crate is not speeding up, where is the energy going? That is, what must be happening to the energy given to the crate as a result of the work done by this 700 N force?
Suppose a horizontal force of 700 N is required to push a 200 kg crate across a warehouse floor at a constant speed of 2.5 m/s. (a) What is the power of this force? (b) If the crate is not speeding up, where is the energy going? That is, what must be happening to the energy given to the crate as a result of the work done by this 700 N force?
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem 6:**
Suppose a horizontal force of 700 N is required to push a 200 kg crate across a warehouse floor at a constant speed of 2.5 m/s.
(a) What is the power of this force?
(b) If the crate is not speeding up, where is the energy going? That is, what must be happening to the energy given to the crate as a result of the work done by this 700 N force?
**Explanation:**
- **Part (a):** The power can be calculated using the formula:
\[ \text{Power} = \text{Force} \times \text{Velocity} \]
For this problem, it translates to:
\[ \text{Power} = 700 \, \text{N} \times 2.5 \, \text{m/s} \]
\[ \text{Power} = 1750 \, \text{W} \]
So, the power of the force is 1750 Watts.
- **Part (b):** Since the crate is not accelerating, the energy input from the force is being used to overcome friction and other resistive forces. This means the energy is converted primarily into thermal energy due to friction between the crate and the floor, and not into kinetic energy since the speed remains constant.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c71d23a-e881-4e7e-8e49-63201aa561a6%2Fa079c8fd-8e48-44e4-814d-c2a3ab274ccc%2F5232if_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem 6:**
Suppose a horizontal force of 700 N is required to push a 200 kg crate across a warehouse floor at a constant speed of 2.5 m/s.
(a) What is the power of this force?
(b) If the crate is not speeding up, where is the energy going? That is, what must be happening to the energy given to the crate as a result of the work done by this 700 N force?
**Explanation:**
- **Part (a):** The power can be calculated using the formula:
\[ \text{Power} = \text{Force} \times \text{Velocity} \]
For this problem, it translates to:
\[ \text{Power} = 700 \, \text{N} \times 2.5 \, \text{m/s} \]
\[ \text{Power} = 1750 \, \text{W} \]
So, the power of the force is 1750 Watts.
- **Part (b):** Since the crate is not accelerating, the energy input from the force is being used to overcome friction and other resistive forces. This means the energy is converted primarily into thermal energy due to friction between the crate and the floor, and not into kinetic energy since the speed remains constant.
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