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Forces using Newtons Second law

  • Thread starter gotpink74
  • Start date
  • #1
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Homework Statement


A 1300 kg car in neutral at the top of a 17° inclined 10 m long driveway slips its parking brake and rolls downward. At what speed will it hit the garage door at the bottom of the incline? Neglect all retarding forces.


Homework Equations


(sin17)
(cos17)
x=vi*t+0.5*a*t^2


The Attempt at a Solution


all of these are wrong. I don't understand how I'm supposed to get the answer
2.86 m/s

3.785 m/s

4.9 m/s

7.928 m/s
 

Answers and Replies

  • #2
PhanthomJay
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Are you familiar with Newton's laws? Draw a sketch of the car and identify the component of the gravity force acting on it parallel to the plane. Then use newton 2 to solve for the acceleration, and one of the kinematic equations to solve for the speed.
 
  • #3
49
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how do you find the acceleration is it just 9.8
 
  • #4
1,039
2
That's the downward acceleration due to gravity. What is the component of that acceleration that is moving the car?
 
  • #5
49
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is the acceleration 0 im really confused
 
  • #6
1,039
2
No, the acceleration is not zero.

You have mg pulling the car down, but the car cant just move down, it has to move down the slope right? So there must be a component of the vector mg that would represent the force on the car in a direction parallel to the slope right?
 
  • #7
49
0
is it 12183N
 
  • #8
49
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is the final answer 9.37
 
  • #9
1,039
2
Show me how you got that solution.
 
  • #10
49
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12183/1300=9.371

i found 12183 using a vector triangle
 
  • #11
1,039
2
Where did 12183 come from, is that 1300(g) where your g is something other than 9.8?

If so, that is only the downward force on the car. You need to find the horizontal component of that vector that is parallel to the hyp of the hill.
 
  • #12
49
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would the speed be 3.785
 
  • #13
785
15
would the speed be 3.785
Be systematic. Resolve the horizontal component of the force first. What is sin 17o ? Draw the diagram please. Then take out the force acting ?
Yes that will be mg sinθ.

With that you can easily take out the speed of car.

Were you guessing ? :biggrin:

Realize that vertical component is itself mass x gravity.
 
Last edited:
  • #14
PhanthomJay
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The component of the gravity force along the incline is mgsin17, not mgcos17. You have to be careful with geometry when calculating angles, and determining vector components of the gravity force , parallel and perpendicular to the incline, epecially when doing these problems for the first time. Now apply Newton 2 along thes direction of the incline, to solve for the acceleration along the direction of the incline, and then use one of the kinematic equations you have now memorized, to solve for the speed at the bottom of the 10 m long incline.
 
  • #15
49
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is the acceleration 380.0
 
  • #16
PhanthomJay
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is the acceleration 380.0
Rather than toss out possible answers, you should show how you are arriving at these figures so that we can point out your errors.

-Show how you arrived at the component of the gravity force acting down the incline.
-Then show how you applied Newton 2 to solve for the acceleration.
-Then show how you used one of the kinematic equations to solve for the velocity at the bottom of the incline.
 

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