Confirm correctness of simple Newton's second law-type problem?

  • Thread starter zero_infinity
  • Start date
In summary, the object's velocity at t=3 is 2 m/s in the \hat{i} direction and 3 m/s in the \hat{j} direction. At t=6 the object has stopped moving in the \hat{j} direction. The object's displacement at t=6 is 3.30 m.
  • #1
zero_infinity
17
0

Homework Statement


at t=0, a 4kg mass is moving at 2 m/s[tex]\hat{i}[/tex] + 3m/s[tex]\hat{j}[/tex]
at t=0, F1 and F2 act upon the object
F1= [1 N + (1/2 N/s)t ][tex]\hat{i}[/tex]
F2= -(1/3 N/s2)t2 [tex]\hat{j}[/tex]

a) what is the object's velocity when t=3?
b) at what time does the object stop moving in [tex]\hat{j}[/tex] direction?
c) what is the object's displacement when t=6?

Homework Equations


F=ma
kinematics

The Attempt at a Solution


just look at the steps and see if I'm doing it right please

[tex]\vec{V}[/tex]0 = 2 m/s[tex]\hat{i}[/tex] + 3m/s[tex]\hat{j}[/tex]
[tex]\vec{F}[/tex](t) = [1 N + (1/2 N/s)t ][tex]\hat{i}[/tex] - (1/3 N/s2)t2 [tex]\hat{j}[/tex]


(a)
[tex]\vec{F}[/tex](t) = m[tex]\vec{a}[/tex]
[tex]\vec{F}[/tex](3)/4 kg = [tex]\vec{a}[/tex]

Use [tex]\vec{V}[/tex]f = [tex]\vec{V}[/tex]0 + [tex]\vec{a}[/tex]t

[tex]\vec{V}[/tex]f = 21/8 [tex]\hat{i}[/tex] - 3/4 [tex]\hat{j}[/tex]





(b)
just look at the [tex]\hat{j}[/tex] component of the vector

[tex]\vec{F}[/tex]j(t)/ 4kg = [tex]\vec{a}[/tex]j
Use [tex]\vec{V}[/tex]f = [tex]\vec{V}[/tex]0 + [tex]\vec{a}[/tex]t

0 = 3m/s[tex]\hat{j}[/tex] - [tex]\vec{a}[/tex]jt

t = 3.30 s





(c)
[tex]\vec{F}[/tex](t)/ 4kg = [tex]\vec{a}[/tex](t)

actually, now when I'm rewriting it, I'm not sure how to integrate [tex]\vec{a}[/tex]

do i integrate the acceleration from 0 to 6? i don't think so...

i integrated acceleration (not accounting for C) and then integrated velocity from 0 to 6 (this is obviously wrong now that I'm looking over it)
 
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  • #2
Find the velocity and acceleration along i and j directions separately.
Using proper kinematic equations find the remaining answers by using vector addition.
 
  • #3
rl.bhat said:
Find the velocity and acceleration along i and j directions separately.
Using proper kinematic equations find the remaining answers by using vector addition.

Hi

which part are you talking about?
 
  • #4
Voi = 2 m/s, F1i = 1N. Find a1i.
Similarly find Voj, F2j and a2J.
Calculate Vfi and Vfj after 3 s.
 

1. What is Newton's second law of motion?

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to its mass. In other words, the greater the force applied to an object, the greater its acceleration will be.

2. How do you confirm the correctness of a simple Newton's second law-type problem?

To confirm the correctness of a simple Newton's second law-type problem, you can follow these steps:

  • Identify the forces acting on the object
  • Determine the net force by adding or subtracting the individual forces
  • Calculate the acceleration using the formula a = F/m, where a is acceleration, F is net force, and m is mass
  • Compare the calculated acceleration with the expected acceleration based on the given problem

3. What are the common mistakes made when solving a Newton's second law-type problem?

Common mistakes when solving a Newton's second law-type problem include:

  • Misinterpreting the given problem and identifying the wrong forces
  • Forgetting to consider all the forces acting on the object
  • Using the wrong formula for calculating acceleration
  • Incorrectly converting units

4. Can Newton's second law be applied to objects with varying mass?

Yes, Newton's second law can be applied to objects with varying mass. The formula a = F/m takes into account the mass of the object, so the acceleration will change accordingly.

5. How does Newton's second law relate to real-life situations?

Newton's second law can be observed in many real-life situations, such as:

  • A car accelerating when the gas pedal is pressed, as the force from the engine increases
  • A ball rolling down a hill, as the force of gravity pulls it downwards
  • A rocket taking off, as the force of the engines propels it upwards

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