A package is at rest on a conveyor belt, which is initially at rest. The belt is started and moves to the right for 1.3 s with a constant acceleration of 2 m/s^2 . the belt then moves with a constant deceleration a2 and comes to a stop after a total displacement of 2.2 m. knowing that the coefficients of friction between the package and the belt are Ms=0.35 and Mk=0.25, determine (a) the deceleration a2 of the belt, (b) the displacement of the package relative to the belt as the belt comes to a stop.
To solve this problem, we need to analyze the motion of the package on the conveyor belt. Let's break it down step by step:
Step 1: Determine the initial velocity of the package
Since the package is initially at rest and the belt is also at rest, the initial velocity (v0) of the package is 0 m/s.
Step 2: Determine the distance traveled during acceleration
Using the equation of motion, s = ut + (1/2)at^2, where s = displacement, u = initial velocity, a = acceleration, and t = time, we can calculate the distance traveled during acceleration:
s1 = (0 m/s)(1.3 s) + (1/2)(2 m/s^2)(1.3 s)^2
s1 = 0 m + (1/2)(2 m/s^2)(1.69 s^2)
s1 = 0 m + 1.69 m
s1 = 1.69 m
Step 3: Determine the distance traveled during deceleration
We know that the total displacement is 2.2 m, and during acceleration, the package traveled 1.69 m. Therefore, during deceleration, the package will travel:
s2 = Total displacement - Distance traveled during acceleration
s2 = 2.2 m - 1.69 m
s2 = 0.51 m
Step 4: Determine the deceleration (a2) of the belt
To find the deceleration (a2), we can use the equation of motion:
v^2 = u^2 + 2as
where v = final velocity (0 m/s), u = initial velocity (from the start of deceleration = 0 m/s), a = acceleration (deceleration), and s = displacement (0.51 m):
0^2 = 0^2 + 2(a2)(0.51 m)
0 = 0 + 2(a2)(0.51 m)
0 = 2(a2)(0.51 m)
0 = 1.02(a2)
a2 = 0 m/s^2
Therefore, the deceleration (a2) of the belt is 0 m/s^2.
Step 5: Determine the displacement of the package relative to the belt
Since the deceleration (a2) of the belt is 0 m/s^2, there is no change in velocity. Therefore, the displacement of the package relative to the belt is equal to the distance traveled during deceleration, which is 0.51 m.
Answer:
(a) The deceleration (a2) of the belt is 0 m/s^2.
(b) The displacement of the package relative to the belt as the belt comes to a stop is 0.51 m.