7. A box of mass m = 80 kg is resting on a horizontal surface. The coefficient of kinetic friction between the box and the surface is 0.300 and the coefficient of static friction is 0.500. Use g = 10/2 Part A: A horizontal pushing force of 300 N is applied to the box. Draw a FBD for the box in this case. No double subscripts. Include a coordinate system.
7. A box of mass m = 80 kg is resting on a horizontal surface. The coefficient of kinetic friction between the box and the surface is 0.300 and the coefficient of static friction is 0.500. Use g = 10/2 Part A: A horizontal pushing force of 300 N is applied to the box. Draw a FBD for the box in this case. No double subscripts. Include a coordinate system.
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![**Question 7**
A box of mass \( m = 80 \, \text{kg} \) is resting on a horizontal surface. The coefficient of kinetic friction between the box and the surface is 0.300, and the coefficient of static friction is 0.500. Use \( g = 10 \, \frac{m}{s^2} \).
**Part A: A horizontal pushing force of 300 N is applied to the box.**
1. **Draw a Free Body Diagram (FBD):**
- Include a coordinate system. (No drawing provided here.)
2. **What is the maximum possible static friction in this case? Show your math:**
The maximum static friction \( f_s \) is given by:
\[
f_s = \mu_s \times N
\]
where \( \mu_s = 0.500 \) is the coefficient of static friction and \( N \) (normal force) is the weight of the box:
\[
N = m \times g = 80 \, \text{kg} \times 10 \, \frac{m}{s^2} = 800 \, \text{N}
\]
Thus,
\[
f_s = 0.500 \times 800 \, \text{N} = 400 \, \text{N}
\]
3. **What is the net force in the x-direction in this case? Explain why:**
The net force is the applied force minus the frictional force. Since the applied force (300 N) is less than the maximum static friction (400 N), the box does not move, and thus the net force is zero.
4. **What is the actual friction in this case?**
The friction is **static** and has a value of 300 Newtons, matching the applied force to prevent motion.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fff53762f-9787-4937-a3ea-4f4cc255b8cf%2F35966c37-4ada-4748-a07f-7ba9b42308fe%2Fjmv2usp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 7**
A box of mass \( m = 80 \, \text{kg} \) is resting on a horizontal surface. The coefficient of kinetic friction between the box and the surface is 0.300, and the coefficient of static friction is 0.500. Use \( g = 10 \, \frac{m}{s^2} \).
**Part A: A horizontal pushing force of 300 N is applied to the box.**
1. **Draw a Free Body Diagram (FBD):**
- Include a coordinate system. (No drawing provided here.)
2. **What is the maximum possible static friction in this case? Show your math:**
The maximum static friction \( f_s \) is given by:
\[
f_s = \mu_s \times N
\]
where \( \mu_s = 0.500 \) is the coefficient of static friction and \( N \) (normal force) is the weight of the box:
\[
N = m \times g = 80 \, \text{kg} \times 10 \, \frac{m}{s^2} = 800 \, \text{N}
\]
Thus,
\[
f_s = 0.500 \times 800 \, \text{N} = 400 \, \text{N}
\]
3. **What is the net force in the x-direction in this case? Explain why:**
The net force is the applied force minus the frictional force. Since the applied force (300 N) is less than the maximum static friction (400 N), the box does not move, and thus the net force is zero.
4. **What is the actual friction in this case?**
The friction is **static** and has a value of 300 Newtons, matching the applied force to prevent motion.
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