Losing Cargo. A box of mass 12.9 rests on the flat floor of a truck. The coefficients of friction between the

Question

box and floor are = 0.190 and = 0.150. The truck stops at a stop sign and then starts to move with an acceleration of 2.22

If the box is a distance 1.79 from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck?
Take the free fall acceleration to be = 9.80
In seconds
How far does the truck travel in this time?
Take the free fall acceleration to be = 9.80 .
In meters

Answer 1

Look at a FBD of the box
First question is does an acceleration of 2.22 cause the box to move?

F=m*a
m*g*0.190=m*a
max a before the box slides
9.8*0.190
=1.862, yup, the box slides

The acceleration of the box is
m*g*0.150=m*a
a=g*0.150
or
1.47 m/s^2
and the acceleration of the truck is
2.22 m/s^2


set
x(t)=1.79+.5*(1.47-2.22)*t^2
and solve for t when x(t)=0
t=2.184

the truck will travel
=.5*2.22*2.184^2

=5.29 meters

j

Answer 2

The initial velocity of the truck = seventy 5.4 km/hr = 20.944 m/sec For the container to start up sliding a pressure F ? 0.32mg is needed, the position m is the mass of the container. This pressure is derived by technique of the reaction pressure of the truck decelerating. even if that is decelerating by technique of -a m/sec/sec, the container will adventure reaction (a ahead) pressure. = ma. For the container no longer to slip ma ? = 0.32mg max. a ? a = 0.32g = 3.139 m/sec² Now utilizing the relation v² - u² = 2as for the truck, with a = - 3.139 m/sec² noting that the truck is composed of halt at s with v =0 , u = 20.944 m/sec- - 20.944² = 2*(-3.139)*s giving s = 20.944²/ (2*3.139) = sixty 9.874 m. The struck ought to provide up in a minimum distance of sixty 9.87 (say 70 m with uniform deceleration ? 3.139m/sec/sec)