The coefficent of static friction between the floor of a truck and a box resting on it is 0.38. The truck is traveling at 79.7 km/hr. What is the least distance in which the truck can stop and ensure that the box does not slide?
2006-12-04 03:35:51 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
79.7 km/hr = 79.7 x 10/36 m/s =22.14 m/s.
Let it stop at a distance S.
The acceleration a = - (22.14) ^2 / 2 S. (Using v^2 -u^2 = 2as).
Force on the box = m a
Frictional force = 0.38 mg.
These two are equal but opposite.
ma = 0.38 mg
a = 0.38 g.
(22.14) ^2 / 2 S = 0.38 x 9.8.
245.09 / S = 3.724
S = 245.09 / 3.724 = 65.8m.
2006-12-04 04:18:44 · answer #1 · answered by Pearlsawme 7 · 0⤊ 0⤋
The box is capable of a deceleration of 9.81 m/sec^2 * .38. Convert the velocity of the truck to meters / second and then determine the distance that corresponds to a deceleration of (9.81)*.38 m/sec^2.
2006-12-04 03:40:57 · answer #2 · answered by www.HaysEngineering.com 4 · 1⤊ 0⤋