College physics homework mathphys man i need ur hwlp

5.9 m/s²

78 N

Explanation:

Draw a free body diagram for each mass.

The mass on the left (M) has a tension force pulling up (T) and a weight force pulling down (Mg).

The mass on the right (m) has a tension force pulling up (T) and a weight force pulling down (mg).

Apply Newton's second law to the mass on the left (remember it accelerates down):

∑F = ma

T − Mg = M (-a)

T − Mg = -Ma

Apply Newton's second law to the mass on the right:

∑F = ma

T − mg = ma

Two equations, two unknowns (T and a).  First, we want to find a.  So start by subtracting the first equation from the second equation.

(T − mg) − (T − Mg) = ma − (-Ma)

T − mg − T + Mg = ma + Ma

Mg − mg = (m + M) a

a = g (M − m) / (m + M)

Given M = 20 kg and m = 5 kg:

a = 9.8 m/s² (20 kg − 5 kg) / (5 kg + 20 kg)

a = 5.88 m/s²

Now plug into either equation to find the tension.

T − mg = ma

T = mg + ma

T = m (g + a)

T = (5 kg) (9.8 m/s² + 5.88 m/s²)

T = 78.4 N

Rounding to two significant figures, the acceleration and tension are 5.9 m/s² and 78 N.

Notice that the radius of the pulley and the distance between the masses were extra information.

e/c, where e is its total energy