A 50-kg box is placed in the bed of a truck.The coefficient of friction between the box and the truck bed is 0.54.If the truck is traveling at 87 km/h (~54 mph)then calculate the minimum distance the truck can stop in without the box sliding into the cab. A)99 m B)1.4 $\times$ 10³ m C)55 m D)1.2 $\times$ 10² m E)2.8 $\times$ 10² m

Question

Quizplus · Accepted Answer

First, we need to calculate the stopping distance of the truck without the box:
v = 87 km/h = 24.17 m/s (conversion: 1 km/h = 0.27777778 m/s)
u = 0 m/s (initial velocity is 0, truck is starting from rest)
a = -0.5*v^2 / d (using the formula v^2 = u^2 + 2ad)
d = -0.5*v^2 / a

We know that friction force (f) = coefficient of friction (μ) * normal force (N).
The normal force is the force exerted by the truck bed on the box and has the same magnitude as the weight of the box: N = mg = 50 kg * 9.81 m/s^2 = 490.5 N
So, the maximum static friction force that can act on the box is f = μN = 0.54 * 490.5 = 264.87 N.

Using Newton’s second law, F = ma (force equals mass times acceleration), we can find the deceleration (a) of the box:
F = ma
F = friction force = 264.87 N
m = mass of the box = 50 kg
a = -F/m

Now, we can use the same formula as before to calculate the stopping distance of the box:
v = 0 m/s (final velocity is 0, the box stops)
u = 24.17 m/s (initial velocity is the same as the truck's)
a = -F/m

d = -0.5*v^2 / a

d = -0.5*(24.17^2) / (-264.87/50) = 55.12 m

Therefore, the minimum distance required to stop the truck without the box sliding into the cab is 55 m.

A 50-Kg Box Is Placed in the Bed of a Truck.The