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dumperdery
Joined: 23 May 2010 Posts: 4 Location: Canada

Posted: Fri May 28, 2010 8:08 pm Post subject: Physics Problem 2  Dynamic Coefficient of Friction 


A 4.1kg box is pushed along a horizontal floor by a
force of magnitude 21 N at an angle θ = 35° with
the horizontal. If the coefficient of kinetic friction
between the block and the floor is 0.20, calculate the acceleration of the box. Be
careful when calculating the normal force.
What do you get when calculating this answer?
I get 3.36m/s^2 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Fri May 28, 2010 10:45 pm Post subject: 


My reasoning for this problem is as follows:
There are two things that need to be calculated before we can answer the question. The first is the normal force acting on the mass as this provides the opposing force of friction.
The normal force is given by:
F_norm = mu*m*g = 0.2*4.1*9.81 = 8.04 N
Next we need to know the horizontal component of the force pushing the mass. This is given by:
F cos(theta) = 21* cos(35) = 17.2 N
The resultant force that accelerates the mass is therefore:
F_res = F  F_norm = 9.16 N
Therefore the acceleration, from Newton's second law F_res = ma
a = 9.16/4.1 = 2.23 ms^2 

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