**Physics Problem: Calculating Frictional Force**A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If the coefficients of friction are:- Static friction coefficient (\(\mu_s\)) = 0.5- Kinetic friction coefficient (\(\mu_k\)) = 0.4Determine the magnitude of the frictional force on the crate.**Options:**- 40 N- 8 N- 12 N- 20 N- 16 N**Solution:**First, calculate the maximum static friction using \( f_s = \mu_s \times N \), where \( N \) is the normal force. In this case, \( N = 40 \, \text{N} \).Thus, \( f_s = 0.5 \times 40 \, \text{N} = 20 \, \text{N} \).Since the applied force (12 N) is less than the maximum static friction (20 N), the crate will not move. Therefore, the frictional force is equal to the applied force, 12 N. The correct answer is **12 N**.

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Description: **Physics Problem: Calculating Frictional Force**A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If the coefficients of friction are:- Static friction coefficient (\(\mu_s\)) = 0.5- Kinetic friction coe

The weight of the box acts in the vertical direction. When the 12N force is applied on the crate,…

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A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If the coefficients of friction are ls = 0.5 and uk = 0.4, the magnitude of the frictional force on the crate is:

A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If the coefficients of friction are ls = 0.5 and uk = 0.4, the magnitude of the frictional force on the crate is:

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Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

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**Physics Problem: Calculating Frictional Force**

A 40-N crate rests on a rough horizontal floor. A 12-N horizontal force is then applied to it. If the coefficients of friction are:

- Static friction coefficient (\(\mu_s\)) = 0.5
- Kinetic friction coefficient (\(\mu_k\)) = 0.4

Determine the magnitude of the frictional force on the crate.

**Options:**
- 40 N
- 8 N
- 12 N
- 20 N
- 16 N

**Solution:**

First, calculate the maximum static friction using \( f_s = \mu_s \times N \), where \( N \) is the normal force. In this case, \( N = 40 \, \text{N} \).

Thus, \( f_s = 0.5 \times 40 \, \text{N} = 20 \, \text{N} \).

Since the applied force (12 N) is less than the maximum static friction (20 N), the crate will not move. Therefore, the frictional force is equal to the applied force, 12 N.

The correct answer is **12 N**.

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