Question

1. a 50 kg mass is initially at rest on a frictionless horizontal surface. a 120 n force is applied to the mass horizontally and to the right. a 2nd force of 122.6 n is applied to the mass horizontally and to the left. these forces are applied for 20 seconds. a. draw a free body diagram of the forces acting on the mass and write net force equations.

Explanation:

Step1: Draw the free - body diagram

• Represent the mass as a rectangle.
• Draw an upward vertical arrow for the normal force \(N\) (since the surface is horizontal and there is no vertical acceleration, \(N = mg\) where \(m = 50\space kg\) and \(g=9.8\space m/s^{2}\), \(N = 50\times9.8=490\space N\)).
• Draw a downward vertical arrow for the weight \(W=mg = 490\space N\).
• Draw a right - ward horizontal arrow for the force \(F_1 = 120\space N\).
• Draw a left - ward horizontal arrow for the force \(F_2=122.6\space N\).

Step2: Write the net - force equations

• In the vertical direction (\(y\) - direction): \(\sum F_y=N - W\). Since there is no vertical motion, \(a_y = 0\). Using Newton's second law \(\sum F_y=ma_y\), we get \(N - W=0\) (because \(a_y = 0\) and \(m

eq0\)).

• In the horizontal direction (\(x\) - direction): \(\sum F_x=F_1 - F_2\). Using Newton's second law \(\sum F_x=ma_x\), where \(m = 50\space kg\), \(F_1 = 120\space N\) and \(F_2 = 122.6\space N\). So, \(ma_x=120 - 122.6\)

Answer:

• Vertical net - force equation: \(N - mg=0\) (where \(N\) is the normal force, \(m = 50\space kg\) and \(g = 9.8\space m/s^{2}\))
• Horizontal net - force equation: \(ma_x=F_1 - F_2\) (where \(m = 50\space kg\), \(F_1 = 120\space N\), \(F_2=122.6\space N\) and \(a_x\) is the horizontal acceleration)