Conservation of Energy block of mass

[SamTsui86]

A small block of mass m1 = 0.600 kg is released from rest at the top of a curved wedge of mass m2 = 4.50 kg, which sits on a frictionless horizontal surface as in Figure P6.59a. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure P6.59b.

What is the height h of the wedge?

Isn't this problem just conservation of energy
I did
KE+PE initial = KE + PE final
0 + mgh = 1/2 mv^2 + 0 and solve for h v is 4.0 m/s
9.8 h = 8
h= .82
but it saids i am wrong, please help

 

[OlderDan]

A small block of mass m1 = 0.600 kg is released from rest at the top of a curved wedge of mass m2 = 4.50 kg, which sits on a frictionless horizontal surface as in Figure P6.59a. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure P6.59b.

What is the height h of the wedge?

Isn't this problem just conservation of energy
I did
KE+PE initial = KE + PE final
0 + mgh = 1/2 mv^2 + 0 and solve for h v is 4.0 m/s
9.8 h = 8
h= .82
but it saids i am wrong, please help

It is not just conservation of energy, and you have not accounted for all of the energy. The wedge is not stationary in this problem. What is conserved?

 

[SamTsui86]

oh i understand now, i have to include the kinetic energy of the wedge in the formula, thank you

 

[OlderDan]

oh i understand now, i have to include the kinetic energy of the wedge in the formula, thank you

And you have to include another conservation principle

 

[SamTsui86]

oh ok, i got the answer now, thank you