How far does a sliding block on an incline go before stopping?

[a18c18]

Homework Statement

A 5.0 kg block slides along a frictionless horizontal surface with a speed of 7.0 m/s. After sliding a distance of 3.0 m, the block makes a smooth transition to a frictionless ramp inclined at an angle of 40° to the horizontal. How far up the ramp does the block slide before coming momentarily to rest?
m

Homework Equations

F=ma
v^2=v(initial)^2+2ad

The Attempt at a Solution

v(initial)=7.0 m/s
F(net)=Fn-Fg; Fn-Fg=ma
Fn=5*9.81*cos 40
Fg=5*9.81
F(net)=-11.4755
a=F(net)/m
a=-2.2951

0=v(initial)^2+2ad
0=7^2+2(-2.2951)d
0=49-4.5902d
d=49/4.5902
d=10.67

I tried this but it was incorrect. I think I might need to account for the 3 meters before the incline in order to find the initial velocity but since the surface was frictionless and there was no given acceleration is there a change in velocity during that time?

 

[Hootenanny]

You might want to reconsider the net force acting on the block when it is on the incline.

 

[a18c18]

Since it is a frictionless surface I didn't think there were any forces other than normal and gravity?

 

[Hootenanny]

Since it is a frictionless surface I didn't think there were any forces other than normal and gravity?

You're correct, the only two forces are the normal force and the weight of the mass. However, what is the net force acting on the mass? In which direction does it act?

 

[a18c18]

Oh okay so it was sin instead of cos thank you very much!

 

[Hootenanny]

Oh okay so it was sin instead of cos thank you very much!

No, what I'm saying is that the normal force has nothing to do with it! The only net force acting on the mass is the component of gravity which acts parallel to the slope, the normal force is irrelevant.