Solving 2kg Fan-Cart Motion: Acceleration, Velocity & Displacement

[jammhawk]

Homework Statement

A 2 kg fan-cart is in linear motion; it experiences 4 N forward force due to the fan.

Assume that the 2kg fan-cart rolls under fan's force, between the weels and track there is 1N frictional force.

1. its acceleration
2. its velocity 3 second after start
3. its velocity 5 second after start
4. its displacement between 3s and 5s
5. its displacement between 0s and 5s
6. its average velocity using the result of 4

Homework Equations

The Attempt at a Solution

Any HELP will be GREAT!

 

[PhanthomJay]

You are going to have to show how you would apply Newton's 2nd law and the general kinematic motion equations.

 

[thomate1]

I would first gather all the relevant information given in the problem statement. The fan-cart has a mass of 2 kg and is experiencing a 4 N forward force due to the fan. Additionally, there is a 1 N frictional force between the wheels and track.

To solve for the acceleration, I would use Newton's second law, F=ma, where F is the net force and m is the mass of the object. In this case, the net force would be the difference between the forward force and the frictional force, which would be 3 N. So, the acceleration would be 3 N / 2 kg, which is equal to 1.5 m/s^2.

To solve for the velocity at 3 seconds after the start, I would use the equation v = u + at, where v is the final velocity, u is the initial velocity (which is assumed to be 0 since the cart starts from rest), a is the acceleration calculated above, and t is the time. So, at 3 seconds, the velocity would be 4.5 m/s.

Similarly, to solve for the velocity at 5 seconds after the start, I would use the same equation, but with t = 5 seconds. So, the velocity at 5 seconds would be 7.5 m/s.

To calculate the displacement between 3 seconds and 5 seconds, I would use the equation s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time. So, the displacement between 3 seconds and 5 seconds would be (4.5 m/s)(2 s) + (1/2)(1.5 m/s^2)(2 s)^2, which is equal to 7.5 m.

The displacement between 0 seconds and 5 seconds is simply the total distance traveled by the cart in 5 seconds. Using the same equation as above, the displacement would be (0 m/s)(5 s) + (1/2)(1.5 m/s^2)(5 s)^2, which is equal to 18.75 m.

To calculate the average velocity using the result from question 4, I would use the equation v = (u + v)/2, where u and v are the initial and final velocities, respectively. So, the