Answer:
s= 0.11 m
Explanation:
Given that
m = 6.6 kg
Initial velocity ,u= 1.51 m/s
Angle ,θ = 26.4°
Coefficient of friction ,μ = 0.57
The final velocity of the block = 0 m/s
s=Distance cover by block coming to rest
Now by using energy conservation
[tex]\left (\mu mgcos\theta+mgsin\theta \right )s =\dfrac{1}{2}mu^2[/tex]
[tex]\left (\mu gcos\theta+gsin\theta\right)s =\dfrac{1}{2}u^2[/tex]
2 s (μ g cosθ + g sinθ) = u²
2 x s x 10 (0.57 x cos26.4° + sin26.4°) = 1.51²
Now by solving above the equation
we find ,s= 0.11 m
Therefore the answer will be 0.11 m.
The distance the block slides before coming to rest is 0.12 m.
The given parameters;
Apply the principle of conservation of energy to determine the distance traveled by the block;
[tex]-mgsin(\theta)d - \mu F_nd = \frac{1}{2} mv^2\\\\-mgsin(\theta)d - \mu mgcos(\theta) d= \frac{1}{2} mv^2\\\\\-md(gsin(\theta) + \mu gcos(\theta))= \frac{1}{2} mv^2\\\\-d(gsin(\theta) + \mu gcos(\theta)) = \frac{1}{2} v^2\\\\-d(9.8\times sin(26.4) \ + 0.57\times9.8 \times cos(26.4)) = \frac{1}{2} \times 1.51^2\\\\-d(4.357 \ + \ 5) = 1.14\\\\d = - \frac{1.14}{9.357} \\\\d = - 0.12 \ m\\\\d = 0.12 \ upwards[/tex]
Thus, the distance the block slides before coming to rest is 0.12 m.
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