How to Find the Final Speed of an Experimental Train?

[xlogit3k]

Homework Statement

In the 1950’s, an experimental train that had
a mass of 22500 kg was powered across a level
track by a jet engine that produced a thrust
of 4.63 × 105 N for a distance of 533 m.
Find the work done on the train.
Answer in units of J.

Homework Equations

The Attempt at a Solution

My attempted solution was to divide the force by the mass but this didnt work.

 

[collinsmark]

My attempted solution was to divide the force by the mass but this didnt work.

Dividing the force by the mass gives you the acceleration, not the work.

Knowing the acceleration, you could use kinematics to find the final velocity and then calculate kinetic energy. But there is a much easier way.

What is the very definition of work? (Hint: it's a function of force and displacement. )

 

[xlogit3k]

Dividing the force by the mass gives you the acceleration, not the work.

Knowing the acceleration, you could use kinematics to find the final velocity and then calculate kinetic energy. But there is a much easier way.

What is the very definition of work? (Hint: it's a function of force and displacement. )

force x distance right?

 

[collinsmark]

force x distance right?

That's right (well, technically the dot product of force and displacement). And the both force and displacement are given in the problem statement.

 

[rwisz]

I would approach this problem by first understanding the concept of work done. Work is defined as the product of force and displacement in the direction of the force. In this case, the force acting on the train is the thrust produced by the jet engine and the displacement is the distance traveled by the train.

Therefore, the work done on the train can be calculated by multiplying the force (4.63 × 105 N) by the displacement (533 m) in the direction of the force. This results in a value of 2.47 × 108 J.

To find the final speed of the train, we can use the equation for kinetic energy, which is defined as ½ * mass * velocity^2. Since the train starts from rest, its initial kinetic energy is zero. Therefore, the work done on the train (2.47 × 108 J) must be equal to the final kinetic energy.

Rearranging the equation, we get:
Final kinetic energy = 2.47 × 108 J = ½ * mass * velocity^2

Solving for velocity, we get:
Velocity = √(2 * 2.47 × 108 J / 22500 kg) = 125.4 m/s

Therefore, the final speed of the train is 125.4 m/s.