Why is this wrong, Spring Problem

  • Thread starter Thread starter Jacob87411
  • Start date Start date
  • Tags Tags
    Spring
AI Thread Summary
The discussion centers on a physics problem involving a block attached to a spring, focusing on calculating the block's speed as it passes through equilibrium. The initial approach incorrectly applies gravitational potential energy instead of elastic potential energy. The correct formula for elastic potential energy is E = 1/2 k Δl², where k is the spring constant and Δl is the displacement from equilibrium. The correct calculation should use this formula to find the speed of the block. Understanding the distinction between gravitational and elastic potential energy is crucial for solving such problems accurately.
Jacob87411
Messages
170
Reaction score
1
A 2.00 kg block is attached to a spring of force constant 590 N/m as in Figure 7.10. The block is pulled 6.00 cm to the right of equilibrium and released from rest.

(a) Find the speed of the block as it passes through equilibrium if the horizontal surface is frictionless.

We can use potential energy here...MGH=1/2MV^2
2GH=V^2
2(9.8)(.06)=V^2
V=1.0844 m/s

Why is this off?
 
Physics news on Phys.org
because it isn't gravitational potential energy... it's elastic potential energy.
 
The elastic potential energy is E=\frac{1}{2}k\Delta l^2[/color]
 

Thread 'Struggling to understand the concept of this question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Newton's law problem: Pushing 2 stacked blocks on a horizontal table
Replies
13
Views
3K
Why can we move the spring with constant speed when we apply a force?
Replies
29
Views
2K
Distribution of Energy when work is done on a system of 2 masses connected by a spring
Replies
17
Views
2K
Spring Problem Involving Variables and Constants Only
Replies
3
Views
1K
Time period of SHM & Energy conservation in pulley-spring-block system
Replies
24
Views
3K
Determine expressions for the following in terms of M, X, D, h and g
Replies
30
Views
1K
Spring and block on an inclined plane
Replies
27
Views
9K
Conservation of Energy: Spring pushing a block up an incline
Replies
15
Views
4K
How to solve a differential equation for a mass-spring oscillator?
Replies
2
Views
2K
Question about springs and rotational/translational energy
Replies
4
Views
1K

Hot Threads

Recent Insights

Back
Top