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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 4.1-kg 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
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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