33) A 5.00-kg object placed on a frictionless, horizontal table is
connected to a string that passes over a pulley and then is fastened
to a hanging 9.00-kg object, as in Figure 8 below Draw free-body
diagrams of both objects. Find the acceleration of the two objects
and the tension in the string
Free-body diagram for the 5.00-kg object:
- Tension force (T) to the right
- Weight force (W) downward
- Normal force (N) upward
- Since the table is frictionless, there are no friction forces
Acceleration of the 5.00-kg object can be found using Newton's Second Law:
ΣF = ma
T - W = ma
T - mg = ma
T = m(a+g)
Free-body diagram for the 9.00-kg object:
- Tension force (T) upward
- Weight force (W) downward
Acceleration of the 9.00-kg object can be found the same way:
ΣF = ma
T - W = ma
T - mg = ma
T = m(a+g)
Since the objects are connected by the same string and the tension in the string is the same throughout, we can set the two tension expressions equal to each other and solve for acceleration:
5(a+g) = 9(a+g)
5a + 5g = 9a + 9g
4a = 4g
a = g
Now, we can substitute the value of acceleration back into one of the tension equations to find the tension in the string:
T = 5(a+g)
T = 5(2g)
T = 10g
Therefore, the acceleration of both objects is equal to the acceleration due to gravity (9.8 m/s^2) and the tension in the string is equal to 98 N.