The figure illustrates a 1.0 kg block being pulled to the right by a 7.0 N force. Despite the force, the block remains stationary. The static and kinetic friction coefficients are 1.0 and 0.5, respectively. The question posed is: How big is the friction force acting on the block?Diagram Explanation:- The image shows a block labeled "1 kg" on a horizontal surface.- A horizontal arrow pointing right indicates a force of 7.0 N being applied to the block via a rope.Given Values:- Coefficient of static friction (\(\mu_s\)): 1.0- Coefficient of kinetic friction (\(\mu_k\)): 0.5Multiple Choice Options:- 0 N- 4.9 N- 7.0 N- 9.8 N- Some other valueThis setup requires calculating the frictional force when the block is stationary, implying the static friction force is balancing the applied force. The static friction force can be calculated using the formula: \[ f_s = \mu_s \cdot N \]where \( N \) is the normal force, equal to the gravitational force on the block (\( N = m \cdot g \), with \( g = 9.8 \, \text{m/s}^2 \)).Calculate \( f_s \) to find the friction force: \[ N = 1.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]\[ f_s = 1.0 \times 9.8 \, \text{N} = 9.8 \, \text{N} \]Since the block is not moving, the static friction force equals the applied force:\[ f = 7.0 \, \text{N} \]

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The figure shows a 1.0 kg block being pulled to the right by a 7.0 N force. The block is not moving. The coefficients of static and kinetic friction are 1.0 and 0 on the block? ON 4.9 N 1 kg Ms=1 Mk=0.5 7.0 N 9.8 N Some other value 7.0 N

The figure shows a 1.0 kg block being pulled to the right by a 7.0 N force. The block is not moving. The coefficients of static and kinetic friction are 1.0 and 0 on the block? ON 4.9 N 1 kg Ms=1 Mk=0.5 7.0 N 9.8 N Some other value 7.0 N

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ISBN:9781305952300

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Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

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Question

The figure illustrates a 1.0 kg block being pulled to the right by a 7.0 N force. Despite the force, the block remains stationary. The static and kinetic friction coefficients are 1.0 and 0.5, respectively. The question posed is: How big is the friction force acting on the block?

Diagram Explanation:
- The image shows a block labeled "1 kg" on a horizontal surface.
- A horizontal arrow pointing right indicates a force of 7.0 N being applied to the block via a rope.

Given Values:
- Coefficient of static friction (\(\mu_s\)): 1.0
- Coefficient of kinetic friction (\(\mu_k\)): 0.5

Multiple Choice Options:
- 0 N
- 4.9 N
- 7.0 N
- 9.8 N
- Some other value

This setup requires calculating the frictional force when the block is stationary, implying the static friction force is balancing the applied force. The static friction force can be calculated using the formula:

\[ f_s = \mu_s \cdot N \]

where \( N \) is the normal force, equal to the gravitational force on the block (\( N = m \cdot g \), with \( g = 9.8 \, \text{m/s}^2 \)).

Calculate \( f_s \) to find the friction force:

\[ N = 1.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9.8 \, \text{N} \]

\[ f_s = 1.0 \times 9.8 \, \text{N} = 9.8 \, \text{N} \]

Since the block is not moving, the static friction force equals the applied force:

\[ f = 7.0 \, \text{N} \]

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