Answer:
1.43 m/s², 10,300 N
Explanation:
First things first is to draw a picture. Fortunately, one was already provided.
Next, we need to draw a free body diagram for each mass.
The 1200 kg mass has two forces acting on it: tension pulling up and weight pulling down.
Similarly, the 900 kg mass has two forces acting on it: tension pulling up and weight pulling down.
Now we apply Newton's second law. Let's start with the 1200 kg. If we say that up is positive, then:
∑F = ma
T - W = M(-a)
T - Mg = -Ma
Notice that I made the acceleration negative. That's because we know that the heavier mass will be accelerating down.
Now we apply Newton's second law to the 900 kg mass:
∑F = ma
T - W = ma
T - mg = ma
Now we have two equations and two unknowns (T and a). Let's solve for the acceleration first. To do that, let's subtract the two equations so we can eliminate T (if you prefer, you can also use substitution instead):
-mg − (-Mg) = ma − (-Ma)
-mg + Mg = ma + Ma
g (M - m) = a (M + m)
a = g (M - m) / (M + m)
Given that M = 1200 kg, m = 900 kg, and g = 10 m/s²:
a = 10 m/s² (1200 kg - 900 kg) / (1200 kg + 900 kg)
a = 10/7 m/s²
a ≈ 1.43 m/s²
Now let's find the tension by plugging our answer into either of the two equations.
T - mg = ma
T - (900 kg) (10 m/s²) = (900 kg) (10/7 m/s²)
T = 72000/7 N
T ≈ 10,300 N
I'm not sure what rounding rules you need to follow, but hopefully this helps.
