Conservation of mechanical energy question

In summary: KE = 0 because the block starts from rest.In summary, a block of mass 2.5 kg is projected up an incline with a spring (k = 2900 N/m) that is compressed by 14 cm. Using energy equations, the maximum distance traveled by the block up the incline can be calculated to be 2.47 m without considering friction.
  • #1
jgray
10
0

Homework Statement



A block of mass 2.5 kg is placed against a compressed spring (k = 2900 N/m) at the bottom of an inclined plane (angle = 28 degrees). When the spring is released the block is projected up the incline and the spring expands by 14 cm to its normal length.
a.Calculate the maximum distance traveled by the block up the incline without friction.
b.Repeat the calculation with friction, taking µk = 0.22.

Homework Equations



E=1/2kx^2
E=1/2mv^2
E=mgy + 1/2kx^2

The Attempt at a Solution



Stuck on part a since I'm not sure its right.. :
E1=E2
mgh=1/2kx^2
(2.5)(9.8)h=1/2(2900)(.14)^2
24.5h=28.42
h=1.16m
So since I have one side of the triangle and the angle, I should be able to find the distance:
d= h/sin28 = 1.16/0.469471562= 2.47 m

Am I on the right track so far?? Thanks for any advise!
 
Physics news on Phys.org
  • #2
jgray said:

Homework Statement



A block of mass 2.5 kg is placed against a compressed spring (k = 2900 N/m) at the bottom of an inclined plane (angle = 28 degrees). When the spring is released the block is projected up the incline and the spring expands by 14 cm to its normal length.
a.Calculate the maximum distance traveled by the block up the incline without friction.
b.Repeat the calculation with friction, taking µk = 0.22.

Homework Equations



E=1/2kx^2
E=1/2mv^2
E=mgy + 1/2kx^2

The Attempt at a Solution



Stuck on part a since I'm not sure its right.. :
E1=E2
mgh=1/2kx^2
(2.5)(9.8)h=1/2(2900)(.14)^2
24.5h=28.42
h=1.16m
So since I have one side of the triangle and the angle, I should be able to find the distance:
d= h/sin28 = 1.16/0.469471562= 2.47 m

Am I on the right track so far?? Thanks for any advise!
That looks like the right result.

Your energy equations are inconsistent.

PESPRING = (1/2)kx2 .

PEGRAVITY = mgh

E = KE + PE
 

What is the conservation of mechanical energy?

The conservation of mechanical energy is a fundamental principle in physics that states that the total amount of mechanical energy in a closed system remains constant over time. This means that energy cannot be created or destroyed, but can only be transferred from one form to another.

What are the two forms of mechanical energy?

The two forms of mechanical energy are kinetic energy and potential energy. Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object has due to its position or state.

How is mechanical energy conserved in a closed system?

In a closed system, mechanical energy is conserved because any energy lost by one object is gained by another object in the system. This exchange of energy can occur through various forms, such as work, heat, or mechanical interactions.

Why is the conservation of mechanical energy important?

The conservation of mechanical energy is important because it allows us to make accurate predictions about the behavior of objects in a closed system. It also helps us understand the relationship between different forms of energy and how they can be converted from one form to another.

Are there any exceptions to the conservation of mechanical energy?

In some cases, mechanical energy may not be conserved due to external factors such as friction, air resistance, or other forms of energy loss. However, in an ideal closed system where these factors are negligible, the conservation of mechanical energy holds true.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
368
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
449
Replies
10
Views
425
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top