Determine its speed after it has slid a distance of 3.00m down the ramp.

[goracheski]

Homework Statement

A 2.00kg mass starts from rest and slides down a frictionless inclined plane that makes an angle of 30 degrees with the horizontal. Determine its speed after it has slid a distance of 3.00m down the ramp.

Homework Equations

I am leaning about Laws of Conservation of Energy
∑E_k=∑E_p
E_k=(1/2)mv²
E_p=mgh
maybe SOH CAH TOA to get the height

The Attempt at a Solution

using a height of 2.5980, from 3cos(30), i put that into my equation but got 7.1396 m/s
when i know the answer is 5.4 m/s?¿¿

 

[nihil666]

Hey!

You said that you're learning about conservation on Energy,
Then i think it could be usefull the theorem of work and kinect energy.
It simply said that the sum of works that acts on a particle is equal to the change in kinect energy.
When you dou a free body diagram you realize that the only force doing work is the horizontal component of weight,( because the normal force is perpendicular to displacement then work is 0) then you know

mgsin30*3(displacement)=1/2mv^2
you solve for v, and you get 5,4 m/s

Tell me if you already learn the theorem, if your not, there's another way to solve the problem.

 

[gneill]

Homework Statement

A 2.00kg mass starts from rest and slides down a frictionless inclined plane that makes an angle of 30 degrees with the horizontal. Determine its speed after it has slid a distance of 3.00m down the ramp.

Homework Equations

I am leaning about Laws of Conservation of Energy
∑E_k=∑E_p
E_k=(1/2)mv²
E_p=mgh
maybe SOH CAH TOA to get the height

The Attempt at a Solution

using a height of 2.5980, from 3cos(30), i put that into my equation but got 7.1396 m/s
when i know the answer is 5.4 m/s?¿¿

You've got the right idea, but take a closer look at how you determined the value for h.

 

[goracheski]

Thank you gneill I realized I used CAH instead of SOH :S

 

[asteriatic]

Thank you for sharing your attempt at solving this problem. It seems like you have a good understanding of the relevant equations and concepts, but there may be an error in your calculations. Here is how I would approach this problem:

First, let's draw a diagram to visualize the situation:

[Insert diagram here]

We can see that the mass starts from rest at the top of the ramp and slides down to a height of 2.5980m, as you correctly calculated. We can use this height to find the potential energy at the top of the ramp:

Ep = mgh
= (2.00kg)(9.8m/s^2)(2.5980m)
= 50.96J

Next, we can use the Law of Conservation of Energy to find the kinetic energy at the bottom of the ramp:

∑Ek = ∑Ep
(1/2)mv^2 = mgh
(1/2)(2.00kg)v^2 = (2.00kg)(9.8m/s^2)(2.5980m)
v^2 = 50.96m^2/s^2
v = √(50.96m^2/s^2)
v = 7.1396m/s

This is the speed of the mass at the bottom of the ramp. However, we are asked to find the speed after it has slid a distance of 3.00m. In order to do this, we need to use the equation for the distance traveled on an inclined plane:

d = h/sinθ
= 2.5980m/sin(30)
= 2.5980m/0.5
= 5.196m

This tells us that the mass has traveled a distance of 5.196m along the ramp. We can now use this distance to find the speed at the bottom of the ramp:

∑Ek = ∑Ep
(1/2)mv^2 = mgh
(1/2)(2.00kg)v^2 = (2.00kg)(9.8m/s^2)(5.196m)
v^2 = 101.568m^2/s^2
v = √(101.568m^2/s^2)
v = 10.078m/s

This is the speed of the mass at the bottom of the ramp after