Recent content by Dakren12

hahah! sweet. i solved the next one as well.. which is pretty much the question in reverse! yay! thanks Doc.

wait a min... the fk here is acting downward huh? ... since we're pushing it upwards... so (20 + 30) - x = 0 x = 50...! hmm... didnt think to apply fk going downwards... haha.

sigh... i think I'm over thinking it... so friction is approx 20 gravity is about 30... the unknown applied force x would i still use f=ma? so something like (20+x)...minus 30 = 0? ... hopeless ..

so Fnet > fk? and fk = mu k * Fn... so... fk = (.5)(mg*cos37)? ... so confusing.

Newton's Second Law...Incline (constant accel) Homework Statement Given: mass of block of 5.0 kg incline angle is 37theta coeff kinetic friction is 0.50 What is the force, directed up the inline, required to move the block at constant speed up the incline? The...