Differential equation problem

Question

You need to determine the best option to push a small cart from 0 to 5 m/s. Two options are offered. The first one implies a mass of 100 kg, a push force of 200 N and a friction coefficient 5 times the velocity (friction force approximate by 5 v). This system will cost 1$ for every second of use. The second one, the mass is 150 kg, the push force is 300 t N (t is the time in second), and the friction is the same as the other option. However, this one only cost 0.5$ per second of use.

1. Which option (with explanation) is the cheapest to reach 5 m/s?
2. What is the velocity for both option after you have spent 300$ on each?
3. Which is the viable option if your employer now ask that the system reaches 50 m/s?

Hints:

1. Use Newton’s second law to build the equations (Velocity in one dimension)
2. You may not have to solve all equations if you can explain your reasoning; be clear and concise.

Answer 1

To determine the cheapest option to reach 5 m/s, let's analyze both options and calculate the cost for each as well as the final velocity after spending $300 on each. We'll also consider the viability of reaching 50 m/s.

Option 1:

- Mass (m1) = 100 kg

- Push force (F1) = 200 N

- Friction force (F_friction1) = 5v N (where v is the velocity in m/s)

- Cost per second (C1) = $1/s

Option 2:

- Mass (m2) = 150 kg

- Push force (F2) = 300t N (where t is the time in seconds)

- Friction force (F_friction2) = 5v N (same as option 1)

- Cost per second (C2) = $0.5/s

We will use Newton's second law to build equations for both options and determine the time it takes to reach 5 m/s.

For Option 1:

Using Newton's second law, the equation of motion is:

F1 - F_friction1 = m1 * a1

Where:

- F1 is the applied force (200 N).

- F_friction1 = 5v N.

- m1 = 100 kg.

- a1 is the acceleration.

At maximum velocity (5 m/s), acceleration a1 is 0:

200 N - 5v N = 100 kg * 0

Solving for v:

5v = 200 N

v = 40 m/s

So, in Option 1, the final velocity after spending $300 is 40 m/s.

The total cost for Option 1 to reach 5 m/s is 300 seconds * $1/second = $300.

For Option 2:

Using Newton's second law, the equation of motion is:

F2 - F_friction2 = m2 * a2

Where:

- F2 = 300t N.

- F_friction2 = 5v N.

- m2 = 150 kg.

- a2 is the acceleration.

At maximum velocity (5 m/s), acceleration a2 is 0:

300t N - 5v N = 150 kg * 0

Solving for v:

5v = 300t N

v = 60t m/s

To reach a velocity of 5 m/s, we can set v = 5 m/s:

5 m/s = 60t

t = 5/60 = 1/12 s

So, it takes 1/12 seconds to reach 5 m/s in Option 2.

The total cost for Option 2 to reach 5 m/s is (1/12) seconds * $0.5/second = $0.0417.

Now, let's consider the viability of reaching 50 m/s:

Option 1 reaches a maximum velocity of 40 m/s, which is less than 50 m/s. Therefore, Option 1 is not a viable option to reach 50 m/s.

Option 2 reaches a maximum velocity of 60t m/s. To reach 50 m/s, we can set 60t = 50 m/s:

60t = 50 m/s

t = 50/60 = 5/6 s

So, it takes 5/6 seconds to reach 50 m/s in Option 2. Therefore, Option 2 is a viable option to reach 50 m/s.

In conclusion:

- Option 1 costs $300 to reach 5 m/s and reaches a maximum velocity of 40 m/s.

- Option 2 costs $0.0417 to reach 5 m/s and is a viable option to reach 50 m/s.

Option 2 is the cheaper and more viable option for reaching 5 m/s and can also be used to reach 50 m/s if needed.