There are two horizontal forces on the 2.0 kg box in the overhead view of Fig. 5-39 but only one (of magnitude F1 = 20 N) is shown. The box moves along the x axis. For each of the following values for the acceleration ax of the box, find the second force in unit-vector notation: (a) 10 m/s2, (b) 20 m/s2, (c) 0, (d) − 10 m/s2, and (e) −20 m/s2.
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The vector sum of F1 and "the second force (F2) " = M a
if a = 0, F1 = -F2
Okay, well what's the procedure when a isn't = to zero?
To find the second force in unit-vector notation for each given value of acceleration (ax), we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
The equation can be written as:
ΣF = m * a
In our case, the net force acting on the box is the sum of the two horizontal forces. Let's denote the second force as F2.
Since we know the value of the first force (F1 = 20 N), we can write the equation as follows:
F2 + F1 = m * a
Let's calculate the second force F2 for each given value of acceleration.
(a) When ax = 10 m/s^2:
F2 + 20 N = 2.0 kg * 10 m/s^2
F2 + 20 N = 20 N
F2 = 0 N
In this case, the second force (F2) is 0 N.
(b) When ax = 20 m/s^2:
F2 + 20 N = 2.0 kg * 20 m/s^2
F2 + 20 N = 40 N
F2 = 40 N - 20 N
F2 = 20 N
In this case, the second force (F2) is 20 N.
(c) When ax = 0 m/s^2:
F2 + 20 N = 2.0 kg * 0 m/s^2
F2 + 20 N = 0 N
F2 = 0 N - 20 N
F2 = -20 N
In this case, the second force (F2) is -20 N.
(d) When ax = -10 m/s^2:
F2 + 20 N = 2.0 kg * (-10 m/s^2)
F2 + 20 N = -20 N
F2 = -20 N - 20 N
F2 = -40 N
In this case, the second force (F2) is -40 N.
(e) When ax = -20 m/s^2:
F2 + 20 N = 2.0 kg * (-20 m/s^2)
F2 + 20 N = -40 N
F2 = -40 N - 20 N
F2 = -60 N
In this case, the second force (F2) is -60 N.
Therefore, the second forces in unit-vector notation for the given values of acceleration are:
(a) F2 = 0 N
(b) F2 = 20 N
(c) F2 = -20 N
(d) F2 = -40 N
(e) F2 = -60 N