Force/Kinematics Problems

  • Thread starter SkiingAlta
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  • #1
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Hey everyone. Struggling a little bit with a force test I have tomorrow. I've tried a few problems, and don't even know where to begin on another. I have the answers, and I would like to see if they are correct. Thank you so much!

1. An object of mass 70 kg is accelerated by a net force of 20 N. What is its acceleration?

My answer is .286 N


2. A speedboat has a mass of 5.0 x 103 kg. It starts from rest and travels 2.0 x 102 m in 6.0 seconds. The boat undergoes uniform acceleration during the 6.0 seconds. What is the net force on the boat?

My answer is 55,550 N


3. A 75 kg girl traveling at a constant velocity on ice skates suddenly experiences an acceleration of -4 m/s2. What is the sum of the forces acting on her?

My answer is (HAH!). I don't even know where to begin. Any tips?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
1,097
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1. An object of mass 70 kg is accelerated by a net force of 20 N. What is its acceleration?

My answer is .286 N
This looks correct.

2. A speedboat has a mass of 5.0 x 103 kg. It starts from rest and travels 2.0 x 102 m in 6.0 seconds. The boat undergoes uniform acceleration during the 6.0 seconds. What is the net force on the boat?

My answer is 55,550 N

I don't think this is correct. Calculate the acceleration then use with Newton's second law (F = ma) to solve for the net force.

HINT: Vf = Vi + at

3. A 75 kg girl traveling at a constant velocity on ice skates suddenly experiences an acceleration of -4 m/s2. What is the sum of the forces acting on her?

My answer is (HAH!). I don't even know where to begin. Any tips?

Hmmm... Sum of the forces acting on her... I think you've got to use Newtons 2nd law. ;)
 
  • #3
19
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So for #2 if I were to do:

Vf=Vi+at
200=0+a6
200=6a
a=33.33333

Then: F=ma
F=5000(33.333)
F= 166,667 N


Right?
 
  • #4
1,097
3
So for #2 if I were to do:

Vf=Vi+at
200=0+a6
200=6a
a=33.33333

Then: F=ma
F=5000(33.333)
F= 166,667 N


Right?

Looks okay to me! Now try number 3!
 
  • #5
19
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And for the latter,

F=ma
F=75(-4)
sumF= -300

Yay?
 
  • #6
1,097
3
F=ma
F=75(-4)
sumF= -300

Yay?

Correct! You don't need to write sumF =, because when you write F = ma you're already implying that:

[tex]\sum F = ma[/tex]
 
  • #7
19
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Awesome. Thanks so much for the help! One last one possibly?

A brick has a mass of 1.2 kg. A force of 5.4 N just begins to move the brick along the floor with a constant velocity. What is the coefficient of static friction?

So I would use f=μn

5.4=μ*11.76

μ=0.46

Would that be right? Thanks so much for your help!
 
  • #8
1,097
3
A brick has a mass of 1.2 kg. A force of 5.4 N just begins to move the brick along the floor with a constant velocity. What is the coefficient of static friction?

So I would use f=μn

5.4=μ*11.76

μ=0.46

Would that be right? Thanks so much for your help!

Correct! Another key observation I want to bring to your attention is that when you have an object moving at constant velocity then,

[tex]\sum F = 0[/tex]

Why's that you ask? Well remember, with constant velocity acceleration is simply 0. Applying Newtons second law,

[tex]\sum F = ma = m(0) = 0[/tex]
 
  • #9
19
0
Oh. That makes sense! Thanks so much for your help!
 

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