1. The problem statement, all variables and given/known data A 45.0 kg box is pulled with a force of 205 N by a rope held at an angle of 46.5° to the horizontal. The velocity of the box increases from 1.00 m/s to 1.50 m/s in 2.50 s. Calculate a) Net force acting horizontally on the box b) Frictional force on the box c) Horizontal component of the applied force d) Co-efficient of kinetic friction between the box and the floor 2. Relevant equations Kinematics equations [mu][Fn] = frictional force 3. The attempt at a solution a) I found acceleration by: change in velocity = 0.5m/s, change in time = 2.50s v --- t = 0.2m/s^2 Ok I found the summation of the X and Y forces: b) Summation of X forces: [(205)(cos46.5) - frictionalF = (45)(0.2)] frictionalF = 132.1N [<---] Summation of Y forces: [Fn + (205)(sin46.5) - (45.0)(0.2) = 0] normalF = 292.74N c) Now, I found the horizontal component of the applied force to be: [205][cos46.5] = 141.1N, but the answer at the back says 293N. Have I done something wrong? d) I am not sure how I am going to get the coefficient.
The horizontal component of the applied force is 205 cos(46.5)= 141.1 N it is correct. The coefficient of friction is obtained from F_{fr}=μ F_{n}. Calculate Fn from the equation for the y components of forces. ehild