A box rests on the (horizontal) back of a truck. The coefficient of static friction between the box and the surface on which it rests is 0.24.
What maximum distance can the truck travel (starting from rest and moving horizontally with constant acceleration) in 3.0 s without having the box slide?
a. 14 m
b. 11 m
c. 19 m
d. 24 m
e. 29 m
You need to use the following equation that relates the distance, speed, acceleration and time such that:
`d = v_0*t + (at^2)/2`
Since the car starts from rest, the initial speed is `v_0 = 0` and the time of 3 s is provided by the problem, hence, substituting these values in equation above yields:
`d = 9a/2`
You need to find the value of acceleration, hence, using the Newton's second law of motion yields:
`m*a = F_f`
`F_f ` represents the friction force
`F_f = mu*N = mu*m*g => m*a = mu*m*g => a = mu*g`
Since the problem provides the coefficient of static friction mu = 0.24 and considering `g = 9.8 m/s^2` yields:
`a = 0.24*9.8 = 2.352 m/s^2`
Substituting `2.352 ` for a in equation`d = 9a/2` yields:
`d = 9a/2 => d = 9*2.352/2 = 10.584 m` ~~ 11m
Hence, evaluating the maximum distance the truck can travel, without having the box slide, yields `d~~11 m` , thus, you may select the answer b.