2. The initially velocity of the box and truck is 60 mph. When the truck brakes such that the deceleration is constant it takes the truck 350 ft to come to rest. During that time the box slides 10 ft and slams into the end of the truck at B. If the coeff of kinetic friction is 0.3, at what speed relative to the truck does the box hit at B? Ans: 5.29 ft/s.

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

this is true

explanation:

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