Question

1. draw a free body diagram for the following situation: a 43 - kg bag of concrete is pushed to the right with 40 n of force. friction resists the motion with 15 n of force while gravity pulls down on the bag with 421.4 n of force.
2. what is the net force acting on the bag of concrete?
3. what is the resulting acceleration of the bag of concrete?

Explanation:

Step1: Calculate net force

Net force in horizontal direction: \(F_{net - x}=F_{applied}-F_{friction}\). Given \(F_{applied} = 40N\) and \(F_{friction}=15N\), so \(F_{net - x}=40 - 15=25N\). In vertical direction, \(F_{net - y}=F_{gravity}-F_{normal}\). Assuming the bag is on a horizontal surface and not accelerating vertically (\(a_y = 0\)), by Newton's second law \(F_{net - y}=ma_y = 0\), so \(F_{normal}=F_{gravity}=421.4N\). The net force on the bag is the horizontal net force since vertical net force is \(0\).

Step2: Calculate acceleration

Using Newton's second law \(F = ma\), where \(F = F_{net - x}\), \(m = 43kg\) and \(F_{net - x}=25N\). Then \(a=\frac{F_{net - x}}{m}\). Substitute \(F_{net - x}=25N\) and \(m = 43kg\) into the formula: \(a=\frac{25}{43}\approx0.581m/s^{2}\)

Answer:

1. The net force acting on the bag of concrete is \(25N\) (in the horizontal direction).
2. The resulting acceleration of the bag of concrete is approximately \(0.581m/s^{2}\) (in the horizontal direction).