Question

A student riding their scooter is pushing on the ground with a force of 100 N. The force of the air is pushing back on the student with a force of 40 N. The total mass of the scooter and the student is 120.0 kg. What is the approximate acceleration of the student?

Answer 1

The approximate acceleration of the student on the scooter is 0.5 m/s² when accounting for the net force and the total mass of the student and scooter.

Step-by-step explanation:

To calculate the approximate acceleration of the student on the scooter, we must apply Newton's second law of motion, which states that the force equals mass times acceleration (F = ma). In this case, the net force acting on the student and scooter can be found by subtracting the backward force of air resistance (40 N) from the forward force exerted by the student (100 N), resulting in a net force of 60 N. Using the formula F = ma and rearranging to solve for acceleration (a), we have: a = F / m, where F is the net force and m is the total mass. Substituting the given values:

a = 60 N / 120.0 kg = 0.5 m/s².

Therefore, the approximate acceleration of the student is 0.5 m/s².

Answer 2

The approximate acceleration of the student on the scooter, after considering the net force resulting from the push and air resistance, is 0.5 m/s².

Step-by-step explanation:

The student's question is concerning the calculation of acceleration for a system consisting of a student on a scooter, using Newton's second law of motion. The force exerted by the student on the ground is given as 100 N, and the opposing force due to air resistance is 40 N. The total mass of the system (student plus scooter) is stated as 120.0 kg.

To find the acceleration, we first determine the net force acting on the student and the scooter by subtracting the force of air resistance from the force exerted by the student: Net Force = (force exerted by the student) - (force of air resistance) = 100 N - 40 N = 60 N.

Using Newton's second law (F = ma), where 'F' is the net force, 'm' is the mass, and 'a' is the acceleration, we can solve for 'a': a = F/m = 60 N / 120.0 kg = 0.5 m/s2.

Therefore, the approximate acceleration of the student is 0.5 m/s2.