Shooting a block up an incline - work energy

In summary, the distance traveled along the incline by a block of mass m after exiting a spring gun at the bottom of the incline and compressing the spring by an amount x_c is L, given by the equation L = [k(x_c)^2 - 2mg(x_c)sin(theta)]/[2mg[(mu)cos(theta) + sin(theta)]], where k is the spring constant, mu is the coefficient of kinetic friction, theta is the angle of the incline, and g is the acceleration due to gravity. Friction is ignored while the block is inside the gun and the uncompressed spring is at the top of the gun.
  • #1
ph123
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A block of mass m is placed in a smooth-bored spring gun at the bottom of the incline so that it compresses the spring by an amount x_c. The spring has spring constant k. The incline makes an angle theta with the horizontal and the coefficient of kinetic friction between the block and the incline is mu. The block is released, exits the muzzle of the gun, and slides up an incline a total distance L.

Find L, the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside the gun. Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance x_c while inside of the gun). Use g for the magnitude of acceleration due to gravity.
Express the distance L in terms of mu, theta, g, x_c, m, and k.


My energy equation is as follows:

0.5k(x_c)^2 = 0.5mv^2 + (mu)mg((x_c) + L)cos(theta)
0.5k(x_c)^2 = (mu)mg((x_c) + L)cos(theta)
[0.5k(x_c)^2]/[(mu)mgcos(theta)] = x_c + L
L = [0.5k(x_c)^2]/[(mu)mgcos(theta)] - x_c

I know you have to add X_c to L to find the total distance the block is moved by the spring, and I really don't understand why this is wrong. Anyone know?
 
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  • #2
It seems to me that you have included friction while the block is in the gun, if I am reading that correctly.
But anyway, this travels up an incline, did you get all the energy associated with your system?
 
  • #3
0.5k(x_c)^2 = mg((x_c) + L)sin(theta) + (mu)mgLcos(theta)

I guess I forgot the y component

But that leaves me with:

L=[k(x_c)^2 - 2mg(x_c)sin(theta)]/[2mg[(mu)cos(theta) + sin(theta)]]

Which seems to me way too complicated an expression.
 
  • #4
ph123 said:
0.5k(x_c)^2 = mg((x_c) + L)sin(theta) + (mu)mgLcos(theta)

I guess I forgot the y component

But that leaves me with:

L=[k(x_c)^2 - 2mg(x_c)sin(theta)]/[2mg[(mu)cos(theta) + sin(theta)]]

Which seems to me way too complicated an expression.
lol, that is not too bad of an expression!
*edit* - the first equation looks okay to me, I didnt check your algebra in the next step.
 

1. How does the height of the incline affect the work energy of shooting a block up?

The height of the incline does not directly affect the work energy of shooting a block up. However, a higher incline may require more work to overcome the force of gravity and reach the top.

2. What is the relationship between the mass of the block and the work energy required to shoot it up an incline?

The work energy required to shoot a block up an incline is directly proportional to the mass of the block. This means that a heavier block will require more work energy to reach the same height as a lighter block.

3. How does the angle of the incline affect the work energy of shooting a block up?

The angle of the incline does not directly affect the work energy of shooting a block up. However, a steeper incline may require more work to overcome the force of gravity and reach the top.

4. Does the speed of the block affect the work energy required to shoot it up an incline?

Yes, the speed of the block does affect the work energy required to shoot it up an incline. The faster the block is moving, the more work energy is needed to overcome the force of gravity and reach the top.

5. Can the work energy of shooting a block up an incline be converted into other forms of energy?

Yes, the work energy of shooting a block up an incline can be converted into other forms of energy such as kinetic energy and potential energy. As the block moves up the incline, it gains potential energy and as it moves down, it gains kinetic energy.

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