A 13,900 kg truck is moving at 23.5 m/s on a mountain road when the brakes are applied. The brakes FAIL!! Fortunately, the driver sees a deceleration (ouch!) ramp a short distance ahead. This is an incline used by out of control trucks to dissipate their KE by converting it to PE in order to stop safely. Since the brakes have failed, you may ignore friction and air resistance.
a) What is the truck’s original kinetic energy?
b) Find the speed of the truck at a point on the ramp that is 12.5 m higher than the point where he entered the ramp.
c) How far above the point where he entered the ramp will the truck be when it comes to rest?
a) KE=mv²/2= 13900•23.5²/2 = 3.83•10⁶ J.
b) KE= PE₁+KE₁
KE₁ = KE-PE = mv²/2 -mgh₁ =
=3.83•10⁶ - 13900•9.8•12.5=
=3.83•10⁶- 1.7•10⁶ =2.13•10⁶ J.
mv₁²/2 = KE₁
v₁ =sqrt{ 2•KE₁/m }= sqrt{2•2.13•10⁶/13900 }=
=17.5 m/s
b)
KE =PE
mv²/2 = mgh
h= mv²/2mg =
=v²/2g = 23.5²/2•9.8 = 28.2 m
To solve these problems, we need to use the concepts of kinetic energy (KE), potential energy (PE), and the work-energy principle.
a) To find the truck's original kinetic energy, we can use the formula: KE = (1/2)mv^2, where m is the mass of the truck and v is its velocity. Plugging in the values, we have:
KE = (1/2)(13,900 kg)(23.5 m/s)^2
KE = 1/2 * 13,900 * 551.75
KE = 3,826,492.5 J
Therefore, the truck's original kinetic energy is 3,826,492.5 J.
b) To find the speed of the truck at a point on the ramp that is 12.5 m higher, we need to relate the change in kinetic energy to the work done on the truck. The work done on the truck is equal to the change in potential energy (PE) plus the change in kinetic energy (KE). The work done can be calculated using the formula: work = force * distance, where force is equal to the weight of the truck and distance is the change in height.
The weight of the truck can be calculated using the formula: weight = mass * gravitational acceleration. The gravitational acceleration can be approximated as 9.8 m/s².
weight = (13,900 kg) * (9.8 m/s^2)
weight = 136,220 N
The work done is equal to the change in potential energy, which is given by: work = mgh, where m is the mass of the truck, g is the gravitational acceleration, and h is the change in height.
work = (13,900 kg)(9.8 m/s^2)(12.5 m)
work = 1,680,125 J
From the work-energy principle, we have:
work = ΔKE + ΔPE
1,680,125 J = ΔKE + ΔPE
Since we're ignoring air resistance and friction, the change in kinetic energy (ΔKE) is equal to the initial kinetic energy (KE). Therefore:
1,680,125 J = 3,826,492.5 J + ΔPE
To find the change in potential energy (ΔPE), we can rearrange the equation:
ΔPE = 1,680,125 J - 3,826,492.5 J
ΔPE = -2,146,367.5 J
Since the height difference (h) is positive, the change in potential energy (ΔPE) is negative. We can now use the formula for potential energy: PE = mgh.
PE = (13,900 kg)(9.8 m/s^2)(12.5 m)
PE = 1,707,125 J
The total potential energy at the higher point is the change in potential energy plus the final potential energy:
PE_total = ΔPE + PE
PE_total = -2,146,367.5 J + 1,707,125 J
PE_total = -439,242.5 J
Now, we can use the total mechanical energy (KE + PE) to find the speed at the higher point. The total mechanical energy is the sum of the final kinetic energy (KE) and the final potential energy (PE_total).
Total mechanical energy = KE + PE_total
Total mechanical energy = 3,826,492.5 J - 439,242.5 J
Total mechanical energy = 3,387,250 J
To find the speed, we can rearrange the formula for kinetic energy:
KE = (1/2)mv^2
v^2 = 2 * KE / m
v^2 = 2 * 3,387,250 J / (13,900 kg)
v = sqrt(2 * 3,387,250 J / 13,900 kg)
v = sqrt(488.55 m^2/s^2)
v ≈ 22.1 m/s
Therefore, the speed of the truck at a point 12.5 m higher on the ramp is approximately 22.1 m/s.
c) To find the height at which the truck will come to rest, we can set the final kinetic energy to zero (since the truck comes to rest):
KE_final = 0
Using the formula for kinetic energy:
KE_final = (1/2)mv^2
0 = (1/2)(13,900 kg)v^2
0 = (1/2)(13,900 kg)(22.1 m/s)^2
Solving for the height:
0 = (1/2)(13,900 kg)(22.1 m/s)^2
0 = (1/2)(13,900 kg)(488.55 m^2/s^2)
0 = 3,826,492.5 J
Therefore, the truck will be at the same height as it entered the ramp when it comes to rest.