What is the normal force on a block at the top of a loop?

In summary: Great job!In summary, a block with a mass of 1.8 kg is placed on a rough inclined plane 8m above the ground. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop with a radius of 2.0 m. The block's speed is observed to be 11 m/s on the floor. The magnitude of the normal force exerted on the block at the top of the loop is found to be 91.26N, which was incorrect. The correct solution involves using the velocity at the bottom of the loop and results in a normal force of 20.7N.
  • #1
AionTelos
3
0

Homework Statement


A block of mass m = 1.8 kg starts at rest on a rough inclined plane a height H = 8m above the ground. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop the loop of radius R = 2.0 m. On the floor the speed of the block is observed to be 11 m/s.

What is the magnitude of the normal force exerted on the block at the top of the loop?

Homework Equations


(Flipping the y axis)
[tex]N + mg = \frac{mv^2}{R}[/tex]

The Attempt at a Solution


[tex]N = \frac{mv^2}{R} - mg[/tex]
[tex]N = \frac{(1.8kg)(11m/s)^2}{2m} - (1.8kg)(9.8m/s^2) = 91.26N[/tex]

The solution says it's 20.7N, I'm not sure where I'm messing up. Though, I feel that maybe I should be using PE=KE?
 
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  • #2
I think I got it, I'm using the velocity at the bottom of the loop instead of at the top, will post the correct solution once I get it.
 
  • #3
For anyone else wondering here was the solution I came too.
[tex]KE_i = PE_f + KE_f[/tex]
[tex]\frac{1}{2}mv_i^2 = mgh + \frac{1}{2}mv_f^2[/tex]
h = -2R
[tex]v_i^2-4gR = v_f^2[/tex]
[tex]N+mg=\frac{mv^2}{R}[/tex]
[tex]N=\frac{m(v_i^2+4gR)}{R}-mg[/tex]
m=1.8kg
v_i=11[itex]\frac{m}{s}[/itex]
g=9.8[itex]\frac{m}{s^2}[/itex]
R=2m
[tex]N=\frac{1.8((11)^2-4(9.8)(2))}{(2)}-(1.8)(9.8)[/tex]

Phew, everything check out?
 
  • #4
Looks right now.
 

Related to What is the normal force on a block at the top of a loop?

1. What is the magnitude at the top of a loop?

The magnitude at the top of a loop refers to the amount of force or acceleration that is acting on an object at the highest point of a circular motion.

2. How is the magnitude at the top of a loop related to centripetal force?

The magnitude at the top of a loop is directly related to the centripetal force, as it is the force that is responsible for maintaining the circular motion and keeping the object from flying off the loop.

3. What factors affect the magnitude at the top of a loop?

The magnitude at the top of a loop can be affected by the speed of the object, the radius of the loop, and the mass of the object.

4. Why is the magnitude at the top of a loop important?

The magnitude at the top of a loop is important because it determines whether an object will successfully complete the loop without falling off. It also affects the amount of force and acceleration experienced by the object.

5. How can the magnitude at the top of a loop be calculated?

The magnitude at the top of a loop can be calculated using the centripetal force formula: F = mv^2/r, where F is the force, m is the mass of the object, v is the velocity, and r is the radius of the loop. It can also be calculated using the conservation of energy principle, taking into account the object's potential and kinetic energy at the top of the loop.

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