What is the numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall?
2 answers:
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Refer to the diagram shown below.
T = the tension in the rope
N = the the normal reaction (the force that the table exerts on the box)
W = the hanging weight
Assume that the pulley is frictionless.
For equilibrium,
T = W
and
T + N = 77
Therefore
N = 77 - W
(a) When W = 30 N,
N = 77 - 30 = 47 n
Answer: 47 N
(b) When W = 60 N,
N = 77 - 60 = 17 N
Answer: 17 N
(c) When W = 90 N
N = 77 - 90 = - 13 N
There is no normal reaction, and the system is no longer in equilibrium.
Instead, the box will be lifted by a force of 13 N, and the box will accelerate upward.
T = the tension in the rope
N = the the normal reaction (the force that the table exerts on the box)
W = the hanging weight
Assume that the pulley is frictionless.
For equilibrium,
T = W
and
T + N = 77
Therefore
N = 77 - W
(a) When W = 30 N,
N = 77 - 30 = 47 n
Answer: 47 N
(b) When W = 60 N,
N = 77 - 60 = 17 N
Answer: 17 N
(c) When W = 90 N
N = 77 - 90 = - 13 N
There is no normal reaction, and the system is no longer in equilibrium.
Instead, the box will be lifted by a force of 13 N, and the box will accelerate upward.
Answer:
16 m
Explanation:
For an object in uniformly accelerated motion (=with constant acceleration), the distance covered by the body can be found by using the following suvat equation:
where
s is the distance covered
u is the initial velocity
t is the time
a is the acceleration
For the body in this problem,
u = 0 (it starts from rest)
is the acceleration
Substituting t = 4 s, we find the distance covered: