Fbd problems using Newtons second law

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To find the coefficient of kinetic friction for a 12.0 kg wooden crate being pushed with a 30.0 N force at constant velocity, recognize that the net force is zero due to Newton's first law. The friction force opposing the motion equals the applied force of 30.0 N. This friction force can be expressed as the product of the coefficient of kinetic friction and the normal force, which is the weight of the crate (mg). Calculate the normal force using the equation mg = 12.0 kg * 9.8 m/s². Finally, solve for the coefficient of kinetic friction by rearranging the equation to find u_k.
agentlxl
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i need help with this problem please help
(If you use a horizontal force of 30.0 N to slide a 12.0kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?)
PLZ HELP AHHHHH ITS DUE TMRW SO PLZ HELP
 
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agentlxl said:
i need help with this problem please help
(If you use a horizontal force of 30.0 N to slide a 12.0kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?)
PLZ HELP AHHHHH ITS DUE TMRW SO PLZ HELP
This is actually Newton 1...sum of forces = 0 when a body is at rest or moving at constant velocity. So 30N to the right must be balanced by the friction force acting left. Calculate the magnitude of the friction force. Then how would you relate this force with u_k and the Normal force?
 
listen i don't know anything my teacher just tells us to do it but doesn't explain so yeah that's my problem but thnx for answering
 
agentlxl said:
listen i don't know anything my teacher just tells us to do it but doesn't explain so yeah that's my problem but thnx for answering
Don't be a quitter! You've got 30N on the crate to the right, and since the block is moving at constant velocity (not accelerating), Newton and Phanthom say that there must be a 30N force acting on it to the left. That's the only way the net force will add to 0, and that force to the left must be the friction force. And since friction force = (friction coefficient)(Normal force), and since Normal force is just the crate's weight, where weight = mg and where g = 9.8m/s/s, solve for the friction coefficient. I'm sure teach told you that. Now come on, chug it out.
 
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