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Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newton’s Laws, and (3) solve the equations of motion. A block starts from rest at a height of 3.9 m on a fixed inclined plane. The acceleration of gravity is 9.8 m/s 2 . 3.8 kg µ = 0.12 31◦ What is the speed of the block at the bottom of the ramp? Answer in units of m/s.

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W0lf93

8 . 55599 m / s.  
Let : h = 8 . 2 m , m = 5 . 1 kg , = 0 . 22 , = 22 , and v f = final speed . The normal force to the inclined plane is N = mg cos . The sum of the forces par- allel to the inclined plane is F net = ma = mg sin - mg cos a = g sin - g cos Since v 2 f = v 2 + 2 ax = 2 ad (1) along the plane, and neglecting the dimension of the block, the distance, d, to the end of the ramp is d sin = h d = h sin (2) therefore v f = r 2 ah sin = r 2 g h (sin - cos ) sin = p 2 g h (1- cot ) (3) = q 2(9 . 8 m / s 2 )(8 . 2 m)[1- (0 . 22)cot22 ] = 8 . 55599 m / s .
onyebuchinnaji

The final velocity of the block at the bottom of the ramp is 7.82 m/s.

The given parameter;

  • height of the block, h = 3.9 m
  • acceleration due to gravity, g = 9.8 m/s²
  • coefficient of friction, [tex]\mu_k[/tex] = 0.12
  • mass of block, m = 3.8 kg

The vertical component of the force is calculated as;

[tex]F_n = mg cos(\theta) \\\\F_k = \mu_k F_n\\\\F_k = \mu_k\ mg cos(\theta)\\\\[/tex]

The horizonal component of the force is calculated as;

[tex]\Sigma F_x = 0\\\\ mg\ sin(\theta) \ - F_k = ma\\\\ mg \ sin(\theta) \ -\mu_k \ mgcos(\theta) = ma\\\\g \ sin(\theta) - \mu_k \ gcos(\theta) = a\\\\ g(sin(\theta) - \mu_kcos(\theta) = a\\\\ 9.8(sin(31) - 0.12\times cos(31)) = a\\\\ 9.8(0.412) = a\\\\ 4.04 \ m/s^2 = a[/tex]

The final velocity of the block at the bottom of the ramp is given as;

[tex]v_f^2 = v_0^2 + 2as\\\\[/tex]

where;

s is the distance along the inclined plane traveled by the block

This distance = hypotenuse side of the triangle formed by plane

[tex]sin(31) = \frac{3.9}{s} \\\\s = \frac{3.9}{sin(31)} \\\\s = 7.57 \ m[/tex]

The final velocity of the block at the bottom of the ramp is calculated as;

[tex]v_f^2 = v_0^2 + 2as\\\\v_f^2 = 0 + 2(4.04)(7.57)\\\\v_f^2 = 61.17\\\\v_f = \sqrt{61.17} \\\\v_f = 7.82 \ m/s[/tex]

Thus, the final velocity of the block at the bottom of the ramp is 7.82 m/s.

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