Speed of a block into a spring

In summary: However, in this case, the initial PE is 0 since the block is released from rest at the height of the spring, which is the reference point for potential energy. So you only need to equate the initial KE to the final PE and KE. In summary, the speed of the block just before it hits the spring can be found by using energy conservation and kinematic equations. The height of the block can be calculated using the gravitational potential energy equation, and the velocity can be found using the kinematic equation for a free fall. It is not necessary to set the sum of initial PE and KE equal to the sum of final PE and KE, as the initial PE is 0 in this case.
  • #1
mikefitz
155
0
A 0.252-kg block is dropped straight downwards onto a vertical spring. The spring constant of the spring is 57 N/m. The block sticks to the spring and the spring compresses 0.13 m before coming to a momentary halt. What is the speed of the block just before it hits the spring?


well the restoring force F is equal to -kx. I need to find out how many seconds the block was in the air prior to coming in contact with the spring and compressing it .13m. This one I'm having a rough time visualizing though, any tips you have would be great...thanks again
 
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  • #2
Hint: use energy conservation to find the height from which the block is thrown. The rest is simple kinematics.
 
  • #3
PE1 + KE1 = PE2 + KE2
.252kg*9.81m/s^2*h + .5*.252*0^2 = .252 * 9.81 * h + .5 * .252 * v^2

I guess if I knew the height I could plug it into the equation and then figure out the value of v^2 - is this correct?
 
  • #4
* ahh - E = mv^2 /2+ kx^2 / 2

once I find total E I can set them equal and solve for v ...
 
  • #5
Note that the gravitational potential energy of the block in the air before its release is mg(h + 0.13). As it falls on the spring, and as the compression in the string equals 0.13, the gravitational potential energy of the block equals zero, while the spring potential energy equals 1/2*k*0.13^2.
 
  • #6
So kx^2 /2 = mgh

57(.13^2) /2 = .252 (9.81)(h)

h=.1948m+.13= .32483m

PE+KE = PEf+KEf

.252*9.81*(.32483) + .5*.252*v^2 = .252*9.81*-.13 + .5*.252*v^2

But, if I work this one out then my v^2 cancels out leaving me with no solution... am i doing something wrong? thanks a lot
 
  • #7
mikefitz said:
So kx^2 /2 = mgh

57(.13^2) /2 = .252 (9.81)(h)

So far, so good, except 57(.13^2) /2 = .252 (9.81)(h + 0.13). Now solve for h.
 
  • #8
57(.13^2) /2 = .252 (9.81)(h + 0.13)

.48165 = .247 (h+.13)
h = .0648 m

.252 * 9.81 * .0648 + 0 = .252 * 9.81 * -.13 + .5 *.252 v^2
.16019=-.3214+.126v^2
v^2=3.822
v=1.955m/s

Hmm - this seems correct buy my book says otherwise

* edit, yeah I'm still unsure about this one...
 
Last edited:
  • #9
* edit, yeah I'm still unsure about this one...
 
  • #10
mikefitz said:
* edit, yeah I'm still unsure about this one...

Your height is correct. Your energy conservation seems wrong. Now you can define the zero-level of potential energy as the point just before the block reaches the spring. So, you've got: mgh = 1/2 mv^2. Or, you can calculate the velocity by simply using the kinematic equation for a free fall by finding the time first, and then calculating the velocity.
 
  • #11
Ok that worked, but I'm a little confused. I thought you had to set the SUM of PE1 & KE1 equal to PE2 and KE2. Thanks
 
  • #12
mikefitz said:
Ok that worked, but I'm a little confused. I thought you had to set the SUM of PE1 & KE1 equal to PE2 and KE2. Thanks

Yes, you have to set the sum of PE1 and KE1 equal to the sum of PE2 and KE2.
 

Related to Speed of a block into a spring

What is the speed of a block when it is released into a spring?

The speed of a block when it is released into a spring depends on the initial conditions such as the mass of the block, the spring constant, and the initial distance between the block and the spring. It can be calculated using the equation: v = √(2kx/m), where v is the speed, k is the spring constant, x is the initial distance, and m is the mass of the block.

How does the mass of the block affect its speed when released into a spring?

The mass of the block affects its speed when released into a spring because it is a factor in the equation for calculating speed (v = √(2kx/m)). A heavier block will have a slower speed compared to a lighter block when released from the same initial distance into the same spring.

What is the relationship between the spring constant and the speed of a block released into the spring?

The spring constant and the speed of a block released into the spring have an inverse relationship. This means that as the spring constant increases, the speed of the block decreases, and vice versa. This relationship is evident in the equation for calculating speed (v = √(2kx/m)), where k is the spring constant.

Can the speed of a block released into a spring be greater than the initial speed?

No, the speed of a block released into a spring cannot be greater than the initial speed. This is because the spring absorbs some of the initial kinetic energy of the block and converts it into potential energy, causing the block to slow down as it compresses the spring.

How does the initial distance between the block and the spring affect the speed of the block when released?

The initial distance between the block and the spring has a direct impact on the speed of the block when released. As the initial distance increases, the speed of the block also increases. This is because a larger initial distance means the block has more potential energy, which is converted into kinetic energy as the block is released and accelerates towards the spring.

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