Question

1. jelene decides to share a drawing she made from the 3rd grade. the locomotive is able to generate an applied force of 4800 n left. assume that there is no friction or air resistance. answer the following:

a. draw a fbd for the system, you may split it into individual objects. 2 marks - c
b. determine the acceleration of the train 3 marks - a
c. determine the tension force between cart 1 and cart 2. 2 marks - a
mlocomotive = 3400 kg
mcart 1 = 450 kg
mcart 2 = 730 kg

Explanation:

Step1: Calculate the total mass of the system

The total mass \(M_{total}=M_{locomotive}+M_{cart1}+M_{cart2}=3400 + 450+730=\boldsymbol{4580\ kg}\)

Step2: Use Newton's second law \(F = Ma\) to find acceleration

Given \(F = 4800\ N\), from \(a=\frac{F}{M}\), we have \(a=\frac{4800}{4580}\approx\boldsymbol{1.05\ m/s^{2}}\)

Step3: Analyze the forces on Cart 2

Let the tension force between Cart 1 and Cart 2 be \(T\). For Cart 2, using \(F = Ma\), \(T = M_{cart2}\times a\). Substituting \(M_{cart2} = 730\ kg\) and \(a\approx1.05\ m/s^{2}\), we get \(T=730\times1.05=\boldsymbol{766.5\ N}\)

Answer:

a. For the Free - Body Diagram (FBD):

• For the locomotive: A force of \(4800\ N\) to the left (applied force), and tension forces \(T_1\) (connecting to Cart 1) to the right. Also, weight (\(mg\)) acting downwards and normal force (\(N\)) acting upwards.
• For Cart 1: Tension force \(T_1\) (from locomotive) to the left and tension force \(T_2\) (connecting to Cart 2) to the right. Weight (\(mg\)) acting downwards and normal force (\(N\)) acting upwards.
• For Cart 2: Tension force \(T_2\) (from Cart 1) to the left. Weight (\(mg\)) acting downwards and normal force (\(N\)) acting upwards.

b. The acceleration of the train is approximately \(1.05\ m/s^{2}\)
c. The tension force between Cart 1 and Cart 2 is approximately \(766.5\ N\)