Question

3.3.2 test (cst): forces
question 7 of 20
this diagram shows two different forces acting on a skateboarder. the combined mass of the skateboard and the person is 81.5 kg. based on this information, what is the acceleration of the skateboarder?
air resistance = 11.40 n
applied force = 52.80
a. 0.51 m/s² to the right
b. 0.51 m/s² to the left
c. 1.94 m/s² to the right
d. 1.94 m/s² to the left

Explanation:

Step1: Calculate the net force

The net force \(F_{net}\) is the difference between the applied force \(F_{applied}\) and the air - resistance \(F_{air}\).
Since the applied force is to the right (\(F_{applied}=52.80\ N\)) and air - resistance is to the left (\(F_{air} = 11.40\ N\)), we use the formula \(F_{net}=F_{applied}-F_{air}\).

$$F_{net}=52.80 - 11.40=41.4\ N$$

The positive value indicates the net force is to the right.

Step2: Use Newton's second law \(F = ma\) to find acceleration

Newton's second law is \(F=ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. We need to solve for \(a\), so the formula becomes \(a=\frac{F_{net}}{m}\).
Given \(m = 81.5\ kg\) and \(F_{net}=41.4\ N\), we substitute the values:

$$a=\frac{41.4}{81.5}=0.51\ m/s^{2}$$

Answer:

A. \(0.51\ m/s^{2}\) to the right