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Newton's Second Law of Motion (problems)

by love_joyously
Tags: motion, newton
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love_joyously
#1
Jun13-05, 08:11 PM
P: 19
Hi I got a few questions that I could not answer about Newton's Second Law of Motion. Can someone please help me?

1) A 1.08 x 10^3 kg car uniformly accelerates for 12.0s from rest.. During this time the car travels 132 m north. What is the net force acting on the car during this acceleration?

2) A 1.20 x 10^3 kg car accelerates uniformly from 5.0m/s east to 12m/s east. During this acceleration the car travels 94m. What is the net force acting on the car during this acceleration?

I have no idea how my teacher got the answer 1.98 x 10^3 N north for Question 1 and 7.6 x 10^2 N east for question 2. Can anyone help me???
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theCandyman
#2
Jun13-05, 08:35 PM
P: 395
1) First [tex]F = ma[/tex], also you know the distance and time and that it started from rest, you can find the acceleration easily using [tex]d=\frac{at^2}{2}[/tex].

2) I am guessing you are having problems finding the acceleration. The acceleration is uniform here, so use [tex]d=\frac{v_1-v_0}{2}t[/tex] to find the time it takes to travel ([tex]v_0[/tex] and [tex]v_1[/tex] are the inital and final velocities respectively) and then using that time [tex]d=\frac{at^2}{2}+v_0t[/tex].

I hope this helps. Do not be discouraged when you cannot solve these, I almost failed physics my freshman year.
love_joyously
#3
Jun13-05, 08:43 PM
P: 19
ok.. i got number 1 but i'm basically stuck on number 2. can u explain it clearly please?

Jameson
#4
Jun13-05, 09:16 PM
P: 787
Newton's Second Law of Motion (problems)

In order to calculate the net force, you need the mass (given) and the acceleration. You will also indicate a direction of the force for this problem, as force is a vector quantity.

So, we all know [tex]F_{net} = ma[/tex]

You have the mass, so let's solve for the acceleration.

Use one of those nice kinematic equations... I'll pick one.

[tex]v_{f}^2 = v_{i}^2 + 2ad[/tex]

You know everything but "a", so use those algebra skills and solve for it.
HallsofIvy
#5
Jun14-05, 05:46 AM
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PF Gold
P: 39,569
The only difference between 1 and 2 is that in 1 the car starts from rest and in 2 it has initial speed 5 m/s.

With constant acceleration, a, from initial speed v0, an object will have speed v= v0+ at after time t and will have gone distance d= v0t+(1/2)at2. (You probably have those equations, or similar ones, in your textbook.)

In problem 2, you know the initial speed was 5 m/s and that the car went 94 m in 12 seconds: 94= 5(12)+ (1/2)a(12)2. Solve that for a and the use F= ma to find the force.
Jameson
#6
Jun14-05, 11:16 AM
P: 787
I would use the kinematic equation I gave because in the above way requires one to solve for time and then solve for acceration, when two steps aren't neccessary. It is quite correct though, and both ways will yield the same answer.
love_joyously
#7
Jun14-05, 06:17 PM
P: 19
thank you! i got the answer now!


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