19 kg box sliding down a frictionless ramp 4.5 m long

[billyghost]

A 19 kg box starts at rest and slides down a frictionless ramp. The length of the ramp is 4.5 m and the height above the ground at the top is 1.3 m. How fast is the box moving at the bottom of the ramp?

Unsure of inclined planes...help with formulas, etc.?

 

[Doc Al]

Think conservation of mechanical energy (KE + PE). (Measure the potential energy from the bottom of the ramp.)

Mechanical Energy (at top of ramp) = Mechanical Energy (at bottom of ramp)

 

[billyghost]

19 x 1.3 x 9.8 ?

 

[Doc Al]

19 x 1.3 x 9.8 ?

That looks like a calculation of the PE at the top of the ramp: PE = mgh. (The units will be Joules.) Which happens to be the total mechanical energy, since it starts from rest. Since energy is conserved, this also equals the KE at the bottom of the ramp. (${KE} = 1/2 m v^2$)

So set the PE at the top (mgh) equal to the KE at the bottom ($1/2 m v^2$) and solve for the speed.

 

[billyghost]

Therefore,velocity should be the square root of 2 x 9.8 x 1.3

 

[Doc Al]

Right. But be sure to give your answer with the proper units.