Question
What is the minimum force required to move a 50 kg box on a surface with a coefficient of kinetic friction of 0.25, and how far will it move in 2 seconds if the force is applied horizontally?
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Answered by
GPT-4o mini
To determine the minimum force required to move a 50 kg box on a surface with a coefficient of kinetic friction of 0.25, we need to calculate the force of kinetic friction acting on the box.
1. **Calculate the normal force (N):**
The normal force for a horizontal surface is equal to the weight of the box, which can be calculated using:
\[
N = m \cdot g
\]
where:
- \( m = 50 \, \text{kg} \) (mass of the box)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
\[
N = 50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 490.5 \, \text{N}
\]
2. **Calculate the force of kinetic friction (F_k):**
The force of kinetic friction can be calculated using:
\[
F_k = \mu_k \cdot N
\]
where \( \mu_k = 0.25 \) (coefficient of kinetic friction).
\[
F_k = 0.25 \times 490.5 \, \text{N} = 122.625 \, \text{N}
\]
3. **Determine the minimum force required to move the box:**
The minimum force (\( F_{\text{min}} \)) needed to overcome the friction is equal to the force of kinetic friction:
\[
F_{\text{min}} = F_k = 122.625 \, \text{N}
\]
Thus, the minimum force required to move the box is approximately **122.63 N**.
Now, let's determine how far the box will move in 2 seconds if this force is applied horizontally.
1. **Calculate the net force (F_net):**
The net force acting on the box after applying the force (\( F \)) is given by:
\[
F_{\text{net}} = F - F_k
\]
If we apply the minimum force (\( F = 122.625 \, \text{N} \)), the net force would be zero because we would only be overcoming the friction force. To find how far it moves with a greater force, let’s assume we apply a slightly larger force, say \( F = 150 \, \text{N} \).
\[
F_{\text{net}} = 150 \, \text{N} - 122.625 \, \text{N} = 27.375 \, \text{N}
\]
2. **Calculate the acceleration (a):**
Using Newton's second law \( F = m \cdot a \), we can solve for \( a \):
\[
a = \frac{F_{\text{net}}}{m} = \frac{27.375 \, \text{N}}{50 \, \text{kg}} = 0.5475 \, \text{m/s}^2
\]
3. **Calculate the distance traveled in 2 seconds:**
Using the kinematic equation:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Assuming the initial velocity \( v_0 = 0 \) and \( t = 2 \) seconds:
\[
d = 0 + \frac{1}{2} \cdot 0.5475 \, \text{m/s}^2 \cdot (2 \, \text{s})^2
\]
\[
d = \frac{1}{2} \cdot 0.5475 \cdot 4 = 1.095 \, \text{m}
\]
Thus, if a force greater than the minimum force is applied, the box will move approximately **1.095 meters** in 2 seconds.
### Summary:
- Minimum force to move the box: **122.63 N**
- Distance moved in 2 seconds (with a greater force): **1.095 m**
1. **Calculate the normal force (N):**
The normal force for a horizontal surface is equal to the weight of the box, which can be calculated using:
\[
N = m \cdot g
\]
where:
- \( m = 50 \, \text{kg} \) (mass of the box)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)
\[
N = 50 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 490.5 \, \text{N}
\]
2. **Calculate the force of kinetic friction (F_k):**
The force of kinetic friction can be calculated using:
\[
F_k = \mu_k \cdot N
\]
where \( \mu_k = 0.25 \) (coefficient of kinetic friction).
\[
F_k = 0.25 \times 490.5 \, \text{N} = 122.625 \, \text{N}
\]
3. **Determine the minimum force required to move the box:**
The minimum force (\( F_{\text{min}} \)) needed to overcome the friction is equal to the force of kinetic friction:
\[
F_{\text{min}} = F_k = 122.625 \, \text{N}
\]
Thus, the minimum force required to move the box is approximately **122.63 N**.
Now, let's determine how far the box will move in 2 seconds if this force is applied horizontally.
1. **Calculate the net force (F_net):**
The net force acting on the box after applying the force (\( F \)) is given by:
\[
F_{\text{net}} = F - F_k
\]
If we apply the minimum force (\( F = 122.625 \, \text{N} \)), the net force would be zero because we would only be overcoming the friction force. To find how far it moves with a greater force, let’s assume we apply a slightly larger force, say \( F = 150 \, \text{N} \).
\[
F_{\text{net}} = 150 \, \text{N} - 122.625 \, \text{N} = 27.375 \, \text{N}
\]
2. **Calculate the acceleration (a):**
Using Newton's second law \( F = m \cdot a \), we can solve for \( a \):
\[
a = \frac{F_{\text{net}}}{m} = \frac{27.375 \, \text{N}}{50 \, \text{kg}} = 0.5475 \, \text{m/s}^2
\]
3. **Calculate the distance traveled in 2 seconds:**
Using the kinematic equation:
\[
d = v_0 t + \frac{1}{2} a t^2
\]
Assuming the initial velocity \( v_0 = 0 \) and \( t = 2 \) seconds:
\[
d = 0 + \frac{1}{2} \cdot 0.5475 \, \text{m/s}^2 \cdot (2 \, \text{s})^2
\]
\[
d = \frac{1}{2} \cdot 0.5475 \cdot 4 = 1.095 \, \text{m}
\]
Thus, if a force greater than the minimum force is applied, the box will move approximately **1.095 meters** in 2 seconds.
### Summary:
- Minimum force to move the box: **122.63 N**
- Distance moved in 2 seconds (with a greater force): **1.095 m**
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