Forces and constant velocity

  • Thread starter macmac410
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  • #1
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Homework Statement


a car with a constant velocity of 1m/s, accidentally hits a huge block with a mass of 300kg. Instead of stopping the driver constantly maintain the speed, pushing the block along the way until the car runs out of fuel. Find the force applied in the block
if the coefficient of friction in the block and surface is 0.22.

Homework Equations


f=ma


m=300kg
a=0 since velocity is constant

u=Ff/Fn
Fn=W=mg

u=Ff/mg
Ff=u(mg)


The Attempt at a Solution


i dont know where to start but it seems that the forced applied on the block was zero since velocity is constant? Am i riGht?
 

Answers and Replies

  • #2
Doc Al
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i dont know where to start but it seems that the forced applied on the block was zero since velocity is constant? Am i riGht?
No. Since the velocity is constant the net force on the block must be zero, not the applied force. What force acts on the block opposing the applied force?
 
  • #3
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No. Since the velocity is constant the net force on the block must be zero, not the applied force. What force acts on the block opposing the applied force?

friction acts on the block opposing the applied force,
since you said that the net force must be zero hence!
Ffriction=Fapplied?

thanks now its getting clearer,
but how is it happend that frictional force is equal to applied force? If the two force are equal then there must be no motion on them?
 
  • #4
Doc Al
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friction acts on the block opposing the applied force,
since you said that the net force must be zero hence!
Ffriction=Fapplied?
Right. (Now you've got to figure out the friction force.)

thanks now its getting clearer,
but how is it happend that frictional force is equal to applied force? If the two force are equal then there must be no motion on them?
No, it means that the block isn't changing its velocity (once it's moving along with the car, that is). I presume that they want the applied force once constant velocity is attained.
 
  • #5
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Thank you very much Doc Al!
now i can find the right answer..
 
Last edited:

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