Two forces are exerted on a hockey puck which initially stopped on frictionless ice. Force 1 is exerted in the positive x direction with magnitude of 10 N and Force 2 is exerted in the positive Y direction with magnitude 4 N. The hockey puck has a mass of 0.5 kg. What is the magnitude and direction of the resulting acceleration?
sqrt (100 + 16) = sqrt 116
= srt (4*29) = 2 sqrt(29)
tan angle above x axis = 4/10
|a| = (1/.5) *2 sqrt 29
= 4 sqrt 29 at same tan^-1(.4)
To find the magnitude and direction of the resulting acceleration, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, this can be expressed as:
a = F_net / m
where "a" is the acceleration, "F_net" is the net force acting on the object, and "m" is the mass of the object.
In this case, we have two forces acting on the hockey puck: Force 1 in the positive x-direction (10 N) and Force 2 in the positive y-direction (4 N).
To find the net force, we need to find the vector sum of these two forces. Since the two forces are perpendicular to each other, we can use the Pythagorean theorem to determine their magnitude:
F_net = sqrt(F_1^2 + F_2^2)
Substituting the given values:
F_net = sqrt((10 N)^2 + (4 N)^2)
= sqrt(100 N^2 + 16 N^2)
= sqrt(116 N^2)
≈ 10.77 N
Now, we can calculate the acceleration:
a = F_net / m
= 10.77 N / 0.5 kg
≈ 21.54 m/s^2
Thus, the resulting acceleration has a magnitude of approximately 21.54 m/s^2.
To find the direction of the resulting acceleration, we can use trigonometry. The angle between the resulting acceleration and the positive x-axis can be determined using the following equation:
θ = arctan(F_2 / F_1)
Substituting the given values:
θ = arctan(4 N / 10 N)
= arctan(0.4)
≈ 21.8 degrees
Therefore, the direction of the resulting acceleration is approximately 21.8 degrees above the positive x-axis.