Solve Physics Problem: Block Moving Up a Vertical Track

In summary, a block of mass 0.640 kg is compressed a distance x against a horizontal spring with a force constant of 450 N/m. After being released, it travels up a frictionless, horizontal surface to the bottom of a vertical circular track with a radius of 1.00 m. At the bottom of the track, the block has a speed of 14.0 m/s and experiences an average frictional force of 7.00 N while sliding up the track. Using the work-energy theorem, it is determined that x = 0.528 m. To find the speed of the block at the top of the track, the same equation is used with the addition of the work done by friction. When solving
  • #1
KD-jay
7
0

Homework Statement


A block of mass 0.640 kg is pushed against a horizontal spring of negligible mass until the spring is compressed a distance x. The force constant of the spring is 450 N/m. When it is released, the block travels along a frictionless, horizontal surface to point B, the bottom of a vertical circular track of radius R = 1.00 m, and continues to move up the track. The speed of the block at the bottom of the track is vB = 14.0 m/s, and the block experiences an average frictional force of 7.00 N while sliding up the track.
http://members.dslextreme.com/users/adiekit/images/Capture.JPG

(a) What is x?
(b) What speed do you predict for the block at the top of the track?

I've gotten (a) but can not figure out (b).

Homework Equations


Σnc=ΔKE + ΔUg + ΔUs

The Attempt at a Solution



(a)What is x?
Σnc=ΔKE + ΔUg + ΔUs
Σnc=(1/2)*m*vf^2 - (1/2)*m*vi^2 + mghf - mghi + (1/2)*k*xf^2 - (1/2)*k*xi^2
Σnc=(1/2)*m*vf^2 - (1/2)*k*xf^2
0=(1/2)*0.640*14^2 - (1/2)*450*xf^2
x=0.528

This is the one I'm having trouble on
(b) What speed do you predict for the block at the top of the track?
I'm assuming that the work done by the friction is F*Δr = 7.00 * Π since the displacement is half the loop. I'm also assuming that the height at the top of the loop is 2 meters.

Σnc=ΔKE + ΔUg + ΔUs
Σnc=(1/2)*m*vf^2 - (1/2)*m*vi^2 + mghf - mghi + (1/2)*k*xf^2 - (1/2)*k*xi^2
Σnc=(1/2)*m*vf^2 - mghf - (1/2)*k*xf^2
-7.00*Π=(1/2)*0.640*vf^2 - 0.640*9.81*2 - (1/2)*450*(0.528)^2
vf=12.9 m/s
 
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  • #2
Looks like you messed up the sign of the gravitational PE in your last equation.
 
  • #3


I would like to commend the student for their efforts in solving this physics problem. It is clear that they have a good understanding of the relevant equations and have correctly calculated the value for x in part (a). However, I would like to offer some suggestions for part (b).

Firstly, it is important to note that the work done by friction is not simply F*Δr, as suggested by the student. The work done by friction is equal to the product of the frictional force and the displacement along the direction of the force. In this case, the displacement is not simply Π, but rather the length of the track that the block travels along.

Secondly, the student has correctly identified that the height at the top of the loop is 2 meters, but they have not taken into account the change in potential energy due to the spring. Remember, the spring is still compressed at the top of the loop, so it will still have some potential energy.

Finally, it is important to remember that energy is conserved in this system, so the total energy at the bottom of the track (kinetic energy + potential energy) will be equal to the total energy at the top of the track. Using this information and the correct equations, the student should be able to solve for the final speed at the top of the track.

Overall, the student has shown a good understanding of the problem and has made a good attempt at solving it. With a few adjustments and taking into account all of the relevant factors, I am confident that the student will be able to accurately calculate the final speed at the top of the track. Keep up the good work!
 

FAQ: Solve Physics Problem: Block Moving Up a Vertical Track

How do I calculate the acceleration of the block on the vertical track?

To calculate the acceleration of the block, you can use the equation a = F/m, where a is the acceleration, F is the net force acting on the block, and m is the mass of the block.

What forces are acting on the block as it moves up the track?

The main forces acting on the block are the force of gravity pulling it downwards and the normal force from the track pushing it upwards. There may also be frictional forces and air resistance depending on the specific situation.

How do I determine the velocity of the block at a certain point on the track?

You can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. You will need to know the initial velocity and acceleration of the block.

Is the work done on the block by the track positive or negative?

The work done on the block by the track is positive because the normal force is in the same direction as the displacement of the block.

How does the angle of the track affect the motion of the block?

The angle of the track affects the motion of the block by changing the components of the gravitational force and the normal force acting on the block. A steeper angle will result in a larger component of the gravitational force pulling the block downwards and a smaller component of the normal force pushing the block upwards, resulting in a faster acceleration.

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