A child pulls a wagon with a force of 33 N by a handle making an angle of 11 degrees with the horizontal. If the wagon has a mass of 5.5 kg, to the nearest hundredth of a m/s2 what is the acceleration of the wagon?
To find the acceleration of the wagon, 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.
First, let's calculate the horizontal component of the force applied by the child. This can be found using the formula: F_horizontal = F_applied * cos(theta), where F_applied is the force applied by the child and theta is the angle the handle makes with the horizontal.
F_horizontal = 33 N * cos(11 degrees)
F_horizontal ≈ 33 N * 0.9816 (since cos(11 degrees) is approximately 0.9816)
F_horizontal ≈ 32.35 N (rounded to two decimal places)
Now, we have the horizontal force acting on the wagon. As there is no vertical force acting on the wagon, we can assume that all the applied force is solely responsible for the acceleration of the wagon in the horizontal direction.
Next, we can use Newton's second law to find the acceleration of the wagon. The formula is: F = m * a, where F is the net force, m is the mass, and a is the acceleration.
Since the horizontal force is acting in the direction of motion, it can be considered as the net force.
32.35 N = 5.5 kg * a
a = 32.35 N / 5.5 kg
a ≈ 5.89 m/s^2 (rounded to two decimal places)
Therefore, the acceleration of the wagon is approximately 5.89 m/s^2 (to the nearest hundredth of a m/s^2).