To find the box's acceleration, we start by calculating the net force acting on the box. The pulling force is 500 N, and we have a frictional force opposing the motion of 100 N.
The net force (\(F_{net}\)) can be calculated as follows:
\[ F_{net} = F_{applied} - F_{friction} \] \[ F_{net} = 500 , \text{N} - 100 , \text{N} = 400 , \text{N} \]
Next, we can use Newton's second law of motion to find the acceleration (\(a\)). The second law states that \(F_{net} = m \cdot a\), where \(m\) is the mass of the box and \(a\) is the acceleration.
Given that the mass (\(m\)) of the box is 40 kg, we can rearrange the equation to solve for \(a\):
\[ a = \frac{F_{net}}{m} = \frac{400 , \text{N}}{40 , \text{kg}} = 10 , \text{m/s}^2 \]
Calculate the distance moved in 7 seconds
Next, we use the equation of motion to find the distance (\(d\)) the box moves in 7 seconds. Since the box starts from rest, we can use the following formula for distance under constant acceleration:
\[ d = v_0 t + \frac{1}{2} a t^2 \]
Where:
- \(v_0\) (initial velocity) = 0 m/s (since it starts from rest)
- \(a\) = \(10 , \text{m/s}^2\)
- \(t\) = \(7 , \text{s}\)
Substituting the values into the equation:
\[ d = 0 \cdot 7 + \frac{1}{2} \cdot 10 \cdot (7^2) \] \[ d = 0 + 5 \cdot 49 \] \[ d = 245 , \text{m} \]
Calculate the velocity after 7 seconds
Now, we need to find the final velocity (\(v\)) after 7 seconds. We can use the following equation for velocity under constant acceleration:
\[ v = v_0 + a t \]
Using the values:
- \(v_0 = 0 , \text{m/s}\)
- \(a = 10 , \text{m/s}^2\)
- \(t = 7 , \text{s}\)
Substituting the values into the equation:
\[ v = 0 + 10 \cdot 7 = 70 , \text{m/s} \]
Summary of Results
- Acceleration: \(10 , \text{m/s}^2\)
- Distance moved in 7 seconds: \(245 , \text{m}\)
- Velocity after 7 seconds: \(70 , \text{m/s}\)
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