Answer:
Speed with which the box hits the truck at B relative to the truck = 5.29 ft/s
Explanation:
First of, we calculate the deceleration of the truck+box setup using the equations of motion.
x = distance tavelled during the deceleration = 350 ft
u = initial velocity of the truck = 60 mph = 88 ft/s
v = final velocity of the truck = 0 m/s
a = ?
v² = u² + 2ax
0² = 88² + 2(a)(350)
700 a = - 88²
a = - 11.04 ft/s²
But this deceleration acts on the crate as a force trying to put the box in motion.
The motion of the box will be due to the net force on the box
Net force on the box = (Force from deceleration of the truck) - (Frictional force)
Net force = ma
Frictional Force = μmg = 0.3 × m × 32.2 = 9.66 m
Force from the deceleration of the truck = m × 11.04 = 11.04 m
ma = 11.04m - 9.66m
a = 11.04 - 9.66 = 1.4 ft/s²
This net acceleration is now responsible for its motion from rest, through the 10 ft that the box moved, to point B to hit the truck.
x = 10 ft
a = 1.4 ft/s²
u = 0 m/s (box starts from rest, relative to the truck)
v = final velocity of the box relative to the truck, before hitting the truck's wall = ?
v² = u² + 2ax
v² = 0² + 2(1.4)(10)
v² = 28
v = 5.29 ft/s
