Question

ASAP PLEASE HELP WITH THIS 1.Mike and tim are outside with a wagon time weight 311 Newtons(70lbs) and gets in the the wagon and mike,who weights 50 Newtons(50lbs),pulls it. As mike pulls it, he accelerates until reaching a constant speed. After stopping, Mike and Tim switch places . Tim now pulls mike in the wagon, accelerating from stop to a constant velocity. Now, Tim may be bigger then mike but mike was riding in the wagon when it had the greatest acceleration during start up? Why? Use Newtons second law of motion to explain.
2. Now Sare comes along, and she is the exact same size as Mike. However, she is even stronger then mike When she pulls mike in the wagon, she pulls with a greater force than when mike pulls her. Now who is in the wagon when it has the greatest acceleration? Please explain, Usung Newton's second law and please answer correctly

Answer 1

Answer:

1. Mike was riding in the wagon when it had the most acceleration because his light weight compared to Tim's weight required the least effort to move

2. Mike

Explanation:

According to Newton's second law of motion, force is proportional to te rate of change of momentum produced

Mathematically, we can write the above law as follows;

F = m × a

Where;

F = The force acting on the object

m = The mass of object in motion

a = The acceleration of the object

1. The given parameters in the question are;

The weight of Tim = 311 Newtons (70 lbs.)

The weight of Mike = 50 Newtons (50 lbs.)

The minimum force required to pull the wagon to constant speed = The weight of the wagon

With the assumption that the wagon has very little weight, we have

Therefore, when Tim gets in the wagon, the force, 'F' applied by Mike to pull the wagon to constant speed = Mass of Tim, m × Acceleration of the cart, a

Given that mass is proportional to weight, we can write;

Force from mike, Tim on the wagon F₁ = 311 N × a₁

Similarly when they switched places, we have;

Force from Tim, Mike on the wagon  = F₂ = 50 N × a₂

Therefore, for the same force, F₁ = F₂ = F, we have;

a₁ = F₁/(311 N) = F/(311 N)

a₂ = F₂/(50 N) = F/(50 N)

By fraction of numbers, F/(50 N) > F/(311 N) > N), therefore, a₂ > a₁

The acceleration of the wagon when Mike was on the wagon will be more than the acceleration of the wagon when Tim gets in the wagon because for the same applied force, the weight of Mike offer less resistance to move

2. Given that Sare and Mike have the same weight of 50 N each let F₃ represent the force with which she pulls Mike in the wagon, and F₁ represent the force with which Mike pulls her while she is on the wagon, we  are also given that F₃ > F₁

By Newton's second law of motion, we have;

a₃ = F₃/(50 N) and a₁ = F₁/(50 N)

From F₃ > F₁, we have;

F₃/(50 N) > F₁/(50 N)

Therefore;

a₃ > a₁

The acceleration of the wagon when Mike is being pulled by Sare, a₃, is greater than the acceleration of the wagon when Sare is pulled by Mike

Therefore, Mike is on the wagon when it has the greatest acceleration.