Physics Help Forum distance !

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Oct 28th 2009, 12:32 PM #1 Senior Member   Join Date: Sep 2009 Location: St. John's, NL Posts: 124 distance ! A 3.62 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 12.4 N at an angle theta = 20.0° above the horizontal, as shown in the Figure. The coefficient of kinetic friction between the block and the floor is 0.07. What distance does the block travel in a time of 3.70 s after it starts moving? hmm Fn= 3.62 * 9.8 = 35.5 N how do i do this problem?
 Oct 28th 2009, 01:19 PM #2 Senior Member   Join Date: Sep 2009 Location: india Posts: 409 17.89 metres? vertical force balance gives mg=N+ F sin theta, so N= mg- F sin theta SO Fric= mew k N= mew k (mg -F sin theta) so net horizontal force= Fcos theta - mew k (mg- F sin theta) net horizontal acc= above found force/ mass use s= 1/2 acc *t^2 (t given=3.7)
 Oct 28th 2009, 06:26 PM #3 Senior Member   Join Date: Sep 2009 Location: St. John's, NL Posts: 124 aha... thank you !!!

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