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Sagot :
Answer:
A) v1 = -0.256 m/s
v2 = 0.128 m/s
B) v1 = -0.0642 m/s
v2 = 0 m/s
C) v1 = v2 = 0 m/s
Explanation:
We are given;
Spring constant; k = 3.85 N/m
Distance compressed; x = 8 cm = 0.08 m
Mass of left block; m1 = 0.25 kg
Mass of right block; m2 = 0.5 kg
A) From conservation of energy;
½kx² = ½m1•v1² + ½m2•(v2)²
Also from conservation of linear momentum, we know that;
m1v1 = m2v2
Now, v2 = m1•v1/m2
Plugging this for v2 in the first equation gives;
½kx² = ½m1•v1² + ½m2•(m1•v1/m2)²
Making v1 the subject, we have;
v1 = √(kx²(m2))/(m1•m2 + (m1)²))
v1 = √[(3.85 × 0.08² × 0.5)/((0.25 × 0.5) + 0.5²))
v1 = 0.256 m/s
This is the left block which is in the negative x direction and so v1 = -0.256 m/s
We saw that v2 = m1•v1/m2
Thus; v2 = (0.25 × 0.256)/0.5
v2 = 0.128 m/s
B) Force exerted by spring is;
F_s = kx = 3.85 × 0.08
F_s = 0.308 N
Normal forces will be calculated for both blocks as;
N1 = m1•g = 0.25 × 9.8 = 2.45 N
N2 = m2•g = 0.5 × 9.8 = 4.9 N
Let's calculate force of static friction for both blocks;
F_s1 = μN1 and F_s2 = μN2
We are given coefficient of friction as
μ = 0.1.
Thus;
F_s1 = 0.1 × 2.45 = 0.245 N
F_s2 = 0.1 × 4.9 = 0.49 N
F_s1 is lesser than F_s. Thus let's calculate the new compression;
x_1 = 0.245/3.85
x_1 = 0.06364 m
Thus, change in compression is;
Δx = x - x_1
Δx = 0.08 - 0.06364
Δx = 0.01636 m
From conservation of energy, since our coefficient of friction is not zero and we have frictional force, then we use the equation;
½k(x)² - F_s1•Δx = ½m1•v1² + ½k(x_1)²
Making v1 the subject, we have;
v_1 = √(k(x² - x_1²) - 2F_s1•Δx)/m1
v_1 = √(3.85(0.08² - 0.06364²) - 2(0.245 × 0.01636)/0.25
v1 = 0.0642 m/s
Since in negative x direction, then
v1 = -0.0642 m/s
F_s2 is greater than F_s. Thus, it means the right side block will not move and velocity is zero. v2 = 0 m/s
C) Coefficient of friction is now 0.463.
Thus;
F_s1 = 0.463 × 2.45 = 1.13435 N
F_s2 = 0.463 × 4.9 = 2.2687 N
They are both greater than F_s and thus no motion in both cases.
So v1 = v2 = 0 m/s
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