Kinetic engery = -1/2mv^2
Work = Fd
Combine:
-1/2mv^2 - Fd = 0
-1/2mv^2 = (0.30*15000*9.81*cos(6))d = 0
Multiply both sides by 2:
v^2 = 2d(0.30*9.81*cos(6))
Solve for d:
d=v^2 / 2(0.30*9.81*cos96))
v = speed:
d = 35^2 / 2*0.30 * 9.81 * cos(6)
d = 1225 / 5.85376
d = 209.27 meters. Round answer as needed.
The length of the ramp required can be determined by using conservation
of energy principle.
The length of the ramp should be approximately 154.97 meters.
Reasons:
Given parameters are;
The angle of inclination of the ramp, θ = 6.0°
Coefficient of friction, μ = 0.30
Mass of the truck, m = 15,000 kg
Speed of the truck, v = 35 m/s
Required;
The length of the ramp to stop the truck
Solution:
From the law of conservation of energy, we have;
Kinetic energy = Work done against friction + Potential energy gained by the truck at height
K.E. = [tex]W_f[/tex] + P.E.
Kinetic energy of the truck, K.E. = [tex]\frac{1}{2} \cdot m \cdot v^2[/tex]
Therefore;
K.E. = [tex]\frac{1}{2} \times 15,000 \times 35^2 = 9,187,500[/tex]
The kinetic energy of the truck, K.E. = 9,187,500 J
Friction force,[tex]F_f[/tex] = m·g·cos(θ)·μ
Therefore;
[tex]F_f[/tex] = 15,000 × 9.81 × cos(6) × 0.30 = 43,903.169071
Friction force,[tex]F_f[/tex] = 43,903.169071 N
Work done against friction = [tex]F_f[/tex] × d
Therefore;
Work done against friction, [tex]W_f[/tex] = 43,903.169071·d
Potential energy gained, P.E. = m·g·h
The height, h = d × sin(6.0°)
∴ P.E. = 15,000 × 9.81 × d × sin(6.0°) = 147150 × d × sin(6.0°)
Which gives;
9,187,500 J = 43,903.169071·d + 147150 × d × sin(6.0°)
[tex]d = \dfrac{9187500}{43,903.169071 + 147150 \times sin(6.0^{\circ})} \approx 154.97[/tex]
The length of the ramp, d ≈ 154.97 m.
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