The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed of the rigis 72 km/h, determine the shortest distance in which the rig can bebrought to a stop if the load is not to shift

50.97 m

Explanation:

m = Mass of truck

= Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h =

s = Displacement

Force applied

Frictional force

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

Since the obect will be decelerating the acceleration will be

From the kinematic equations we have

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

no

sodium is solid at rt

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j