Net force = F = 60 - 10 = 50 Newtons
F = m a
50 = 25 a
a = 2 m/s^2
v = a t = 2 t (assuming v =zero at t = 0)
x = (1/2) a t^2 = t^2
10 = (1/2)(2) t^2
so t = sqrt (10)
speed at 10 meters = 2 sqrt 10 = ANSWER final speed
so average speed = sqrt (10)
work done= F d = 50*10 = 500 Joules
(1/2) m v^2 = 500
(1/2)(25) v^2 = 500
v^2 = 1000/25 = 40
v = sqrt( 4*10) = 2 sqrt (10) [ whew, same answer ]
A man accelerates a 25 kg box by pushing it with a force of 60 N over a distance of 10.0 m on a surface that provides a frictional force of 10 N
Determine the velocity of the box after the man stops pushing in two ways
a) by calculating net force,the acceleration and then the average speed
b by cacualating net work done on the box and then considering the changes in energy of the box
1 answer
1 answer