Calculate the magnitude of the tension force in the rope

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SUMMARY

The discussion focuses on calculating the tension force in a rope used to pull a 1050-kg car with an acceleration of 1.50 m/s² over 12 seconds. The correct magnitude of the tension force is 1575 N, derived from the formula F=ma, where F is the force, m is the mass, and a is the acceleration. Additionally, the second part of the problem involves calculating the braking force required to stop the car in 5 seconds, which necessitates determining the initial velocity and subsequent deceleration.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Basic knowledge of kinematics, including acceleration and velocity
  • Ability to perform calculations involving mass, force, and acceleration
  • Familiarity with the concept of constant acceleration
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  • Learn how to apply Newton's laws in various physics problems
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psruler
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Can anyone help me on two problems of my hw?

1) A tow rope is used to pull a 1050-kg along a level road, initially accelerating it from rest at 1.50m/s^2 for a duration of 12.0seconds. Calculate the magnitude of the tension force in the rope during this acceleration.(ignore friction in the problem)

My answer was 394N but i want to make sure if that was correct.

2) As a stoplight is approached, the person riding in the towed car applies the brakes and brings it to a stop in an additional 5.00s. Calculate the magnitude of the force of the brakes on the car, assuming the brakes are applied evenly so that the deceleration is constant.
 
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Regarding question 1:

This is clearly an "F=ma" question. A mass of 1050 KG and an acceleration 1.5 m/s^2should give a force of at least 1500N

Question 2:
There's a bunch of ways to do this one. One of the easiers ways is to use v=at to find the starting velocity, and then the force in a similar fashion. Clearly the force here is larger than the force for question 1.
 
Ok, I got the initial velocity for ques.2 and where do I go from there?
 
The change in velocity is equal to the acceleration multiplied by the time. You know the change in velocity, and the time, so that should give you the acceleration.
 
wouldn't that give me an acceleration of 1.5 again?
 
Ok i got it. thanks NateTG for your help!
 

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