A box is sliding with a constant speed of 3.00 m/s in the +x-direction on a hori
ID: 1333425 • Letter: A
Question
A box is sliding with a constant speed of 3.00 m/s in the +x-direction on a horizontal, frictionless surface. At x = 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between x = 0 and x = 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m, it is 0.400.
What is the x-coordinate of the point where the box comes to rest?
How much time does it take the box to come to rest after it first encounters the rough patch at x = 0?
Explanation / Answer
Solution: Part (a)
let the box has mass m. Let us set m = 1 kg (unit mass)
The initial speed of the box v = 3 m/s
During the travel from x = 0m to 2m, the coefficient of kinetic friction is k1= 0.2 and afterward coefficient of kinetic friction is k2=0.4
Frictional force f = k*N = k*mg and the work done by the frinctinal force is W = k*mg*d
where k = coeficient of kinetic friction, N = normal force on the box from the surface which is equal to mg (weight)
The kinetic energy of the box is
K = (1/2)mv2 -------------------------------(1)
Upon encountering the rough patch, when the box comes to rest then all of the kinetic energy is lost by in doing work against the frictional forces.
The work done during box’s travel from x = 0m to 2 m is
W1 = k1*mg*d1 --------------------------------------------------(2)
If we take d1 = 2m then
W1 = = 0.2*mg*2
Now work done during the box’s travel from x = 2m to d2 where box stops
W2 = k2*mg*d2 ----------------------------------------------(3)
From conservation of mechanical energy
K = W1 + W2
(1/2)mv2 = k1*mg*d1 + k2*mg*d2
(1/2)v2 = k1*g*d1 + k2*g*d2
(1/2)(3m/s)2 = 0.2*(9.81m/s2)* d1 + 0.4*(9.81m/s2)* d2
4.5 m2/s2 = 0.2*(9.81m/s2)* d1 + 0.4*(9.81m/s2)* d2
Now consider d1= 2m then 0.2*(9.81m/s2)* (2m) = 3.924 m2/s2 which is less than 4.5m2/s2. It means that the box does not stop at x = 2m but still has kinetic energy to cross x = 2m.
4.5 m2/s2 = 0.2*(9.81m/s2)*(2m) + 0.4*(9.81m/s2)* d2
4.5 m2/s2 - 0.2*(9.81m/s2)*(2m) = 0.4*(9.81m/s2)* d2
0.4*(9.81m/s2)* d2 = 0.576 m2/s2
d2 = (0.576 m2/s2)/( 0.4*9.81m/s2)
d2 = 0.1468 m.
Thus the box comes to rest at x = d1 + d2 = 2m + 0.1468 m
x = 2.1468 m
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Part (b)
During the travel from x = 0m to 2m, the coefficient of kinetic friction is k1= 0.2 and afterward coefficient of kinetic friction is k2=0.4
Thus the deceleration that box is undergoing is from x= 0m to 2m is a1 can be found by
-m*a1 = k1*mg (since frictional force is in opposite direction of motion)
a1 = -k1*g
a1 = -0.2*(9.81m/s2)
a1 = -1.962 m/s2
Now time t1 taken by the box to cover this distance can be found by
d1 = u*t1 + (1/2)a1*t12
2m = (3m/s)*t1 – (1/2)*(-1.962 m/s2)*t12
0.981* t12 + 3* t1 - 2 =0
This is the quadratic equation with solution which can be solved to get
t1 = 0.5630 s or t1 = -3.6211s
Neglecting negative term, we have t1 = 0.5630 s -------------------------------(4)
Speed of the box at x = 2
vmid = u + at1 = 3m/s + (-1.962 m/s2)*( 0.5630s) = 1.8954 m/s
Now the deceleration that box is undergoing is from x= 2m to 2.1468 m is a2 can be found by
-m*a2 = k2*mg
a2 = -k2*g
a2 = -0.4*(9.81m/s2)
a2 = -3.924 m/s2
time t2 taken by the box to cover this distance d2 = 0.1468m can be found by
d2 = vmid*t2 + (1/2)a2*t22
0.1469m = (1.8954 m/s)*t2 – (1/2)*(-3.924 m/s2)*t22
1.962*t22 + 1.8954*t2 – 0.1469 =0
This is the quadratic equation with solution which can be solved to get
t2 = 0.0721 s or t2 = -1.0381 s
Neglecting negative term we have t2 = 0.0721 s
Hence total time taken by box to come to rest is
T = t1 +t2 = 0.5630 s +0.0721 s
T = 0.6351 s
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