Two men are carrying a uniform ladder 6 m long with a mass of 40 kg. if one of the men lifts with 150 N at one end of the ladder, a) where (in terms of distance to the end where the first man lifts), and b) how much force must the second man lift?

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

The answer is frictional force