Answer to Question #151535 in Quantum Mechanics for Kelsey Smith

Question

A 50.0 N force is applied at an angle of 30.0 degrees north of east to a 5.78 kg box. Assuming there is a coefficient of kinetic friction of 0.389 between the table and the box, what is the time it takes to travel 3.00 meters?

Answer 1

$s=at^2/2\to t=\sqrt{\frac{2s}{a}}$

$F=ma\to F\cos30°-fN=ma$

$N=mg-50\sin30°=5.78\cdot9.8-50\cdot\sin30°=31.6(N)$

$a=\frac{F\cos30°-fN}{m}=\frac{50\cdot\cos30°-0.389\cdot 31.6}{5.78}=5.36(m/s^2)$

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2\cdot 3}{5.36}}=1.1(s)$