Question

What force will be needed to move the block with 0.76 N frictional force, 0.15 coefficient of kinetic, and mass of 0.6kg up the inclined plane raised at an angle of 30 degrees so that it will attain a velocity of 1.5 m/s after 3 sec.?

Answer 1

$\vec F=m\vec a$

$F_x=F-F_f-mg\sin30°=ma\to F=ma+mg\sin30°+F_f$

$F_y=N-mg\cos30°=0\to N=mg\cos30°$

$F_f=\mu N=\mu\cdot mg\cos30°$

$v=at\to a=v/t=1.5/3=0.5\ (m/s^2)$

$F=ma+mg\sin30°+F_f=0.6\cdot0.5+0.6\cdot 9.8\cdot \sin30°+0.15\cdot0.6\cdot 9.8\cos30°=$

$=4\ (N)$ . Answer