27.
a)In the 1950’s, an experimental train that had a mass of 32800 kg was powered across a level track by a jet engine that produced a thrust of
4.37× 10^5 N for a distance of 465 m.Find the work done on the train.
Answer in units of J.
b)Find the change in kinetic energy.
Answer in units of J.
c)Find the final kinetic energy of the train if it started from rest.
Answer in units of J.
d)Find the final speed of the train assuming no friction.
Answer in units of m/s.
a) The work done on the train can be calculated using the equation:
Work = Force × Distance
In this case, the force acting on the train is the thrust produced by the jet engine, which is 4.37 × 10^5 N, and the distance covered by the train is 465 m. Plugging in the values, we get:
Work = (4.37 × 10^5 N) × (465 m)
= 2.03 × 10^8 J
So, the work done on the train is 2.03 × 10^8 J.
b) The change in kinetic energy can be calculated using the equation:
Change in Kinetic Energy = Work
Since the work done on the train is the same as the change in kinetic energy, the answer is also 2.03 × 10^8 J.
c) If the train started from rest, its initial kinetic energy would be zero. So, the final kinetic energy would be equal to the change in kinetic energy, which is 2.03 × 10^8 J.
d) Assuming no friction, we can use the equation for kinetic energy:
Kinetic Energy = (1/2) × mass × (velocity)^2
Since the initial kinetic energy is zero, the final kinetic energy is equal to the change in kinetic energy, which is 2.03 × 10^8 J. We can rearrange the equation to solve for velocity:
(1/2) × mass × (velocity)^2 = 2.03 × 10^8 J
Solving for velocity, we find:
(velocity)^2 = (2 × 2.03 × 10^8 J) / mass
Taking the square root of both sides:
velocity ≈ √[(2 × 2.03 × 10^8 J) / 32800 kg]
≈ √(1.22 × 10^4 m^2/s^2)
≈ 110.56 m/s
So, the final speed of the train, assuming no friction, is approximately 110.56 m/s.
To solve these problems, we can use the following formulas:
a) Work done (W) = Force (F) x Distance (d)
b) Change in kinetic energy (ΔKE) = Work done (W)
c) Final kinetic energy (KE) = 1/2 x mass (m) x velocity^2 (v^2)
d) Final speed (v) = √(2 x ΔKE / mass)
Given values:
Mass (m) = 32800 kg
Force (F) = 4.37 x 10^5 N
Distance (d) = 465 m
a) Work done (W) = Force (F) x Distance (d)
W = (4.37 x 10^5 N) x (465 m)
W ≈ 2.029 x 10^8 J
b) Change in kinetic energy (ΔKE) = Work done (W)
ΔKE ≈ 2.029 x 10^8 J
c) Initial kinetic energy (KE_initial) = 0 (since the train started from rest)
Final kinetic energy (KE) = KE_initial + ΔKE
KE ≈ 0 + 2.029 x 10^8 J
KE ≈ 2.029 x 10^8 J
d) Final speed (v) = √(2 x ΔKE / mass)
v = √(2 x 2.029 x 10^8 J / 32800 kg)
v ≈ √(12396341.463) m/s
v ≈ 3518.78 m/s
Therefore:
a) The work done on the train is approximately 2.029 x 10^8 J.
b) The change in kinetic energy is approximately 2.029 x 10^8 J.
c) The final kinetic energy of the train is approximately 2.029 x 10^8 J.
d) The final speed of the train, assuming no friction, is approximately 3518.78 m/s.
To solve this problem, we'll use several basic principles of physics.
a) The work done on an object is equal to the force applied on the object multiplied by the distance over which the force is exerted. In this case, the force is the thrust produced by the jet engine, and the distance is the distance traveled by the train. So, the work done on the train can be calculated by multiplying the force and the distance:
Work = Force × Distance
Given:
Force = 4.37 × 10^5 N
Distance = 465 m
Plugging in these values into the equation, we can find the work done:
Work = (4.37 × 10^5 N) × (465 m)
Work = 2.02845 × 10^8 J
Therefore, the work done on the train is approximately 202,845,000 J.
b) The change in kinetic energy of an object is equal to the work done on it. In this case, the work done on the train will be equal to the change in its kinetic energy:
Change in Kinetic Energy = Work
So, the change in kinetic energy is also approximately 202,845,000 J.
c) To find the final kinetic energy, we need to know the initial kinetic energy of the train. Given that the train started from rest, the initial kinetic energy is zero. The final kinetic energy is therefore the sum of the initial kinetic energy and the change in kinetic energy:
Final Kinetic Energy = Initial Kinetic Energy + Change in Kinetic Energy
Final Kinetic Energy = 0 + 202,845,000 J
Final Kinetic Energy = 202,845,000 J
Therefore, the final kinetic energy of the train is approximately 202,845,000 J.
d) The final speed of an object can be calculated using the equation:
Final Speed = √(2 × Final Kinetic Energy / Mass)
Given:
Final Kinetic Energy = 202,845,000 J
Mass = 32,800 kg
Plugging in the values, we can find the final speed:
Final Speed = √(2 × 202,845,000 J / 32,800 kg)
Final Speed = √(12,285 m²/s²)
Final Speed = 110.89 m/s
Therefore, the final speed of the train assuming no friction is approximately 110.89 m/s.