Answer:
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.
