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The 2000 kg truck rolls down hill reaching a speed of 30 m/s just before hitting a tree. The vehicle stops 0.72 seconds after first making contact with the tree. What is the average force applied by the tree?

physics
Guest Mar 12, 2015

Best Answer 

 #1
avatar+92221 
+5

Please can someone check this.

 

I do not know so I am going to guess. :)

F=mass* acceleration           m=2000kg

 

v=0, u=30m/s, t=0.72sec,  a=?

v=u+at

0=30+a*0.72

-30/0.72=a

a=-41.6 repeater  m/s^2

 

F=2000*41.66666

F=83333   kgm/s^2 = 83,333 Newtons     (to the nearest Newton)

Melody  Mar 13, 2015
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1+0 Answers

 #1
avatar+92221 
+5
Best Answer

Please can someone check this.

 

I do not know so I am going to guess. :)

F=mass* acceleration           m=2000kg

 

v=0, u=30m/s, t=0.72sec,  a=?

v=u+at

0=30+a*0.72

-30/0.72=a

a=-41.6 repeater  m/s^2

 

F=2000*41.66666

F=83333   kgm/s^2 = 83,333 Newtons     (to the nearest Newton)

Melody  Mar 13, 2015

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