Question

7 rock a weighs 200 n and rock b weighs 400 n. they are then both placed on a horizontal floor and pushed with a 1000 n force. which rock will accelerate more horizontally? a rock a b rock b c they both have the same acceleration. d neither rock will accelerate.

Explanation:

Step1: Calculate mass of each rock

Using \(F = mg\) (\(g = 10m/s^{2}\)), for rock A: \(m_{A}=\frac{F_{A}}{g}=\frac{200N}{10m/s^{2}} = 20kg\). For rock B: \(m_{B}=\frac{F_{B}}{g}=\frac{400N}{10m/s^{2}}=40kg\).

Step2: Calculate acceleration using \(F = ma\)

Assume same pushing force \(F = 1000N\) (net force, ignoring friction as not given). For rock A: \(a_{A}=\frac{F}{m_{A}}=\frac{1000N}{20kg}=50m/s^{2}\). For rock B: \(a_{B}=\frac{F}{m_{B}}=\frac{1000N}{40kg} = 25m/s^{2}\). But wait, if we consider that weight \(W=mg\), and assume same coefficient of friction \(\mu\), frictional force \(f=\mu N=\mu mg\). Net force \(F_{net}=F - f=F-\mu mg\). Acceleration \(a=\frac{F_{net}}{m}=\frac{F}{m}-\mu g\). Since \(F = 1000N\), \(W_{A} = 200N\), \(W_{B}=400N\). If we assume \(F\) is the net force (no friction mentioned in problem), \(a=\frac{F}{m}\). \(m\) is proportional to \(W\) (\(W = mg\)). \(a_{A}=\frac{F}{\frac{W_{A}}{g}}=\frac{Fg}{W_{A}}\), \(a_{B}=\frac{Fg}{W_{B}}\). Substituting \(F = 1000N\), \(g = 10m/s^{2}\), \(W_{A}=200N\), \(W_{B}=400N\). \(a_{A}=\frac{1000\times10}{200}=50m/s^{2}\), \(a_{B}=\frac{1000\times10}{400}=25m/s^{2}\). But if we assume that the problem has an error (maybe force is same as weight units confusion), no - if we consider standard Newton's second law \(F = ma\), and \(W=mg\). Rock A has smaller mass (\(m=\frac{W}{g}\)), so for same applied force (net force), \(a=\frac{F}{m}\), smaller \(m\) gives larger \(a\).

Answer:

A. Rock A