The coefficients of static and kinetic friction between a 43 kg box and a horizontal surface are 0.72 and 0.20, respectively.
(a) What is the acceleration of the object if a 135 N horizontal force is applied to the box?
(b) What is the acceleration if the applied force is 480 N
To find the acceleration of the object, we need to consider the forces acting on it and apply Newton's second law of motion.
(a) When a 135 N horizontal force is applied to the box:
1. Determine the force of friction (static friction) acting on the box using the formula:
\(F_{\text{friction}} = \mu_{\text{static}} \cdot F_{\text{normal}}\)
where \(F_{\text{normal}}\) is the normal force acting on the box.
As the box is on a horizontal surface, the normal force is equal to the weight of the box, which is given by:
\(F_{\text{normal}} = m \cdot g\)
where \(m\) is the mass of the box and \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).
Hence,
\(F_{\text{normal}} = 43 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 421.4 \, \text{N}\)
Now, substituting the values into the formula for static friction:
\(F_{\text{friction}} = 0.72 \cdot 421.4 = 303.808 \, \text{N}\)
2. Determine the net force acting on the box by subtracting the force of friction from the applied force:
\(F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 135 - 303.808 = - 168.808 \, \text{N}\)
The negative sign indicates that the net force is acting in the opposite direction of the applied force.
3. Now, apply Newton's second law of motion to find the acceleration:
\(F_{\text{net}} = m \cdot a\)
Substituting the known values:
\(- 168.808 = 43 \cdot a\)
Solving for \(a\):
\(a = \frac{- 168.808}{43} = -3.919 \, \text{m/s}^2\)
The negative sign indicates that the object is experiencing deceleration.
Therefore, the acceleration of the object is approximately \(-3.919 \, \text{m/s}^2\).
(b) When a 480 N horizontal force is applied to the box:
1. Determine the force of friction (kinetic friction) acting on the box using the formula:
\(F_{\text{friction}} = \mu_{\text{kinetic}} \cdot F_{\text{normal}}\)
where \(F_{\text{normal}}\) is the normal force acting on the box (equal to its weight).
As before, the normal force is \(m \cdot g = 43 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 421.4 \, \text{N}\).
Now, substituting the values into the formula for kinetic friction:
\(F_{\text{friction}} = 0.20 \cdot 421.4 = 84.28 \, \text{N}\)
2. Determine the net force acting on the box by subtracting the force of friction from the applied force:
\(F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 480 - 84.28 = 395.72 \, \text{N}\)
3. Apply Newton's second law of motion to find the acceleration:
\(F_{\text{net}} = m \cdot a\)
Substituting the known values:
\(395.72 = 43 \cdot a\)
Solving for \(a\):
\(a = \frac{395.72}{43} = 9.194 \, \text{m/s}^2\)
Therefore, the acceleration of the object is approximately \(9.194 \, \text{m/s}^2\).