A problem about spring constant

In summary, a plane weighing 220 kN (25 tons) lands on an aircraft carrier at a speed of 64.0 m/s (141 mi/h). The arresting cables bring the plane to a stop in a distance of 84.0 m. To calculate the work done by the cables, the formula Work = 1/2(mv^2) was used, resulting in a value of 450.56. However, the mass of the plane was not taken into account, leading to an incorrect answer. After converting the weight into kilograms, the correct answer of 20,000 kg was obtained. The force exerted by the cables was then found to be 5.36 Newtons, which is too low
  • #1
katie7la
1
0
A plane weighing 220 kN (25 tons) lands on an aircraft carrier. The plane is moving horizontally at 64.0 m/s (141 mi/h) when its tailhook grabs hold of the arresting cables. The cables bring the plane to a stop in a distance of 84.0 m.
(a) How much work is done on the plane by the arresting cables?
(b) What is the force (assumed constant) exerted on the plane by the cables?

(a)
For this problem I used
Work = K where K =1/2(mv^2)
k=1/2((220)(64^2))
k = 450.56

(b) work = F*x
450.56 = F*84
F=5.36


However, I got it wrong. What am I doing wrong?
 
Physics news on Phys.org
  • #2
katie7la said:
(a)
For this problem I used
Work = K where K =1/2(mv^2)
k=1/2((220)(64^2))
k = 450.56
What's the mass of the plane? Redo this calculation, taking care with units.

(b) work = F*x
450.56 = F*84
F=5.36
Your method is OK, but your answer from part a is way, way off. (Realize that a force of 5.36 Newtons is about a pound of force--way too low to be realistic.)
 
  • #3
Sorry, but I'm still having trouble. The plane weighs 220kN (25 tons).
So should i convert the tons into kilograms? 1 ton = 1000kg so it would be 25000 kilograms??
and then from there on do i use the equation...
Work = K where K =1/2(mv^2)
 
  • #4
Why do you have two different IDs?
kbyws37 said:
The plane weighs 220kN (25 tons).
So should i convert the tons into kilograms? 1 ton = 1000kg so it would be 25000 kilograms??
That's a metric ton; instead, convert Newton's to kg via: weight = mg.
and then from there on do i use the equation...
Work = K where K =1/2(mv^2)
Yes.
 

What is spring constant?

Spring constant, also known as force constant, is a measure of the stiffness of a spring or other elastic object. It is defined as the force required to stretch or compress the spring by one unit of length.

How is spring constant calculated?

Spring constant can be calculated by dividing the applied force by the resulting displacement. It can also be calculated by dividing the elastic potential energy stored in the spring by the displacement.

What factors affect spring constant?

The factors that affect spring constant include the material and thickness of the spring, the number of coils, and the diameter of the coils. Temperature and external forces can also affect the spring constant.

Why is spring constant important?

Spring constant is important because it determines the amount of force required to stretch or compress a spring. It is also used in various equations to calculate the behavior of springs in mechanical systems.

How does spring constant affect the motion of an object attached to a spring?

Spring constant affects the motion of an object attached to a spring by determining how much force is needed to displace the object from its equilibrium position and how quickly it will return to that position. A higher spring constant will result in a faster oscillation and a greater force needed to stretch or compress the spring.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
236
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Introductory Physics Homework Help
Replies
29
Views
822
  • Introductory Physics Homework Help
Replies
3
Views
308
  • Introductory Physics Homework Help
Replies
10
Views
808
  • Introductory Physics Homework Help
Replies
17
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
3K
Back
Top