An inclined plane is 518 cm long and sits on a horizontal table. One end is lifted by a block 71.5 cm high. Assume the potential energy is zero on the bottom surface of the incline.

NOTE: Be careful of units!

a) If a 807 g mass is placed on the surface of the incline, 343 cm up the incline, find the potential energy of the mass at this point.

b) Suppose now that the 807 g mass is released from rest at the very top of the incline, and it slides down the frictionless surface. Find the speed of the mass at the very bottom of the incline.

## Answers (1)

a) E = mgh

m = 807g * (1kg/1000g) = 0.807kg

g = 9.8 m/s^2

h can be found from similar triangles because only the vertical height off of the table matters.

h / 343 = 71.5/518

h = 47.3 cm * (1 m/100cm) = 0.473m

therefore,

E = .807kg * 9.8 m/s^2 *.473m = 3.74 J

b) E = mgh at the top of the incline (velocity = 0), h = 71.5 cm. At the bottom, all potential energy is kinetic energy so E = .5mv^2. Energy is conserved since there is no friction so

E = mgh = .5*mv^2

or

v = sqrt(2gh)

substitute in

v = sqrt(2*9.8m/s^2 * .715m)

v=3.74 m/s