Conservation of energy problem on an igloo

[ap.p.b.studen]

Homework Statement

Jin is sitting on top of a hemispherical, frictionless igloo of radius 2.40 meters. His friend pushes him, giving him an initial speed. Jin slides along the igloo and loses contact with it after he has traveled 1.60 meters along the surface. What was his initial speed?

Homework Equations

KEi+PEi=KEf+PEf

Ei=Ef

F=macos(angle)

Work=KE+PE

Work=Fx(delta x)

S=(angle)(radius)

PE=mgh

KE=.5(m)(v^2)

The Attempt at a Solution

I drew a picture of a semicircle and used the S=theta(r) equationand trignometry to find that the hieghts (2.40m initial and 1.89m when leaves igloo). I used the conservation of energy to find that the velocity when Jin leaves the igloo is 3.16m/s. How would I find the initial velocity?

 

[paul232]

How did you find 3.16? Kinetic in the initial condition is not zero.. You are trying to find the velocity.. How could it be zero?

Also think about what happens at the exact moment it leaves the semicircle and what kind of motion it executes.. You didnt write one equation you need.

Hope it helps

 

[kerimek]

To find the initial velocity, we can use the equation KE = 1/2mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity. We can also use the equation Work = KE + PE, where Work is the work done on Jin by his friend, KE is the kinetic energy, and PE is the potential energy. Since the igloo is frictionless, there is no work done by friction, so the work done on Jin is equal to the change in potential energy. Therefore, we can set Work = PE and solve for the initial velocity.

Work = PE = mgh = (m)(9.8)(1.89) = 18.522mJ

KE = 1/2mv^2

18.522mJ = 1/2(m)v^2

v = √(37.044mJ/m) = 6.08m/s

Therefore, Jin's initial speed was 6.08m/s when his friend pushed him.