Asked by Ana
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
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
Answers
Answered by
Damon
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 ]
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 ]