(a) figure 1 shows a mas of 1200 kg hung by a massless cable supported by a counterweight of mass 900 kg. you may assume the gravitational acceleration to be 10 m/s2.

determine the magnitude of the acceleration (in m/s2) of the 1200 kg mass and also the magnitude of the tension (in n) in the cable. show your workings clearly.

me bcoz i don't really understand how to do without a proper explanation behind the working. i'm quite weak in this subject, i m sorry : (

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.

the most reliable source of information.is wikipedia