Conservation of mechanical energy problem

In summary, the equation for range as a function of initial speed and angle can be found by solving the equation for V0 and theta. If you use that value for the angle, then the equation provides the longest range for a given speed.
  • #1
BrainMan
279
2
Problem: the masses of the javelin, discus, and shot are .8 kg, 2.0 kg, and 7.2 kg, respectively, and record throws in the track events using these objects are about 89 m, 69 m, and 21 m, respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of 2 m. The answer is (a) 349 j, 676 j, 741 j (b) 175n , 338 n, 371n

Relevant equations: KEi+PEi=KE+PEf
Or 1/2mv^2+mgy= 1/2mv^2+mgy

Attempt: I tried to use some kind of substitution to find v and y but I only had one equation so I couldn't use the substitution method.
 
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  • #2
To have a projectile reach a given range, how can it be set in motion? Is there any freedom in choosing the parameters of the initial motion? How can that freedom be used to minimise the initial kinetic energy?
 
  • #3
BrainMan said:
Problem: the masses of the javelin, discus, and shot are .8 kg, 2.0 kg, and 7.2 kg, respectively, and record throws in the track events using these objects are about 89 m, 69 m, and 21 m, respectively. Neglecting air resistance, (a) calculate the minimum initial kinetic energies that would produce these throws and (b) estimate the average force exerted on each object during the throw, assuming the force acts over a distance of 2 m. The answer is (a) 349 j, 676 j, 741 j (b) 175n , 338 n, 371n

Relevant equations: KEi+PEi=KE+PEf
Or 1/2mv^2+mgy= 1/2mv^2+mgy

Attempt: I tried to use some kind of substitution to find v and y but I only had one equation so I couldn't use the substitution method.
For a given initial velocity, what is the launch angle that produces the longest range (longest horizontal distance of travel before hitting the level ground)? Use that angle when using the formula for initial velocity and range.
AM
 
  • #4
The launch angle is not given.
 
  • #5
BrainMan said:
The launch angle is not given.
Write out the equation for the Range as a function of initial speed and angle. At what angle is the range maximum (i.e. keeping velocity constant)? If you use that value for the angle, then the equation provides the longest range for a given speed. They give you the range. Work out the speed.

AM
 
  • #6
I used the equation
ImageUploadedByPhysics Forums1398872549.449364.jpg
. I used this to try to find the initial velocity and then used the initial velocity to try to find the kinetic energy when the potential energy is zero thus finding the energy needed for the system. I got the wrong answer. Is the formula I derived not correct?
 
  • #7
X represents the x distance at the top of the projectiles motion.
 
Last edited:
  • #8
Equation for trajectory:

$$X = \frac{V_{0}^2\sin(2\theta)}{g}$$

I think they want you to assume your angle ##\theta## is 45 degrees, because that is the angle at which a maximum trajectory will occur.

You are given the trajectories but not the initial velocity, so that is what you want to solve the above equation for.

After that, I am going to assume you know the equations for work and energy. Since they are asking for the average force over a given distance, the standard ##W = Fd## equation can be used.
 
  • #9
The way you wrote that equation is unclear. Could you send me a better representation?
 
  • #10
BrainMan said:
The way you wrote that equation is unclear. Could you send me a better representation?

Umm, not sure how else to write it. If you were to read it out, you'd say that the trajectory of a projectile is equal to the initial velocity squared multiplied by the sine of 2 times the launch angle theta, divided by g, the acceleration of gravity.

##X## = trajectory
##V_{0}## = initial launch velocity
##\theta## = initial launch angle
##g## = acceleration of gravity
 
  • #11
Ok I get it. Thanks!
 

1. What is the conservation of mechanical energy problem?

The conservation of mechanical energy problem is a principle in physics that states that the total amount of mechanical energy in a closed system remains constant over time. This means that energy cannot be created or destroyed, but can only be transferred between different forms (such as potential and kinetic energy).

2. How is the conservation of mechanical energy problem applied?

The conservation of mechanical energy problem is applied in various situations, such as when calculating the motion of a pendulum or a roller coaster. It is also used in fields such as engineering and mechanics to analyze the energy transfers and transformations in mechanical systems.

3. What are the key factors that affect the conservation of mechanical energy?

The key factors that affect the conservation of mechanical energy are potential energy, kinetic energy, and the presence of external forces such as friction. Any changes in these factors can affect the total amount of mechanical energy in a system.

4. Can the conservation of mechanical energy be violated?

No, the conservation of mechanical energy is a fundamental principle in physics and cannot be violated. While energy can be transferred between forms, the total amount of mechanical energy in a closed system will always remain constant.

5. What are some real-life examples of the conservation of mechanical energy problem?

Some real-life examples of the conservation of mechanical energy problem include a swinging pendulum, a bouncing ball, and a roller coaster ride. In each of these situations, the total amount of mechanical energy remains constant, even as the energy is transferred between different forms.

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