To initiate motion, a 50 kg steel box on a steel floor requires an applied force calculated using the equation \( F_{\text{applied}} = \mu_k \cdot F_{\text{normal}} \), yielding approximately 294 Newtons with a assumed coefficient of kinetic friction (\( \mu_k = 0.6 \)).
To determine the force required to start the steel box moving across the steel floor, you can use the equation of motion:
\[ F_{\text{applied}} = \mu_k \cdot F_{\text{normal}} \]
Where:
- \( F_{\text{applied}} \) is the applied force (what we want to find).
- \( \mu_k \) is the coefficient of kinetic friction between the box and the floor.
- \( F_{\text{normal}} \) is the normal force exerted on the box, which is equal to the weight of the box (\( F_{\text{normal}} = m \cdot g \), where \( m \) is the mass of the box and \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \)).
First, let's calculate the normal force:
\[ F_{\text{normal}} = m \cdot g \]
\[ F_{\text{normal}} = 50 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \]
\[ F_{\text{normal}} = 490 \, \text{N} \]
Now, you need to know the coefficient of kinetic friction (\( \mu_k \)). This value depends on the nature of the surfaces in contact (steel on steel, in this case). The coefficient of kinetic friction for steel on steel is typically around 0.6, but this can vary.
Let's assume \( \mu_k = 0.6 \):
\[ F_{\text{applied}} = 0.6 \cdot 490 \, \text{N} \]
\[ F_{\text{applied}} = 294 \, \text{N} \]
So, with the assumed coefficient of kinetic friction, approximately 294 Newtons of applied force is needed to start the 50 kg steel box moving across the steel floor. Keep in mind that this value can change if the coefficient of kinetic friction is different.