physics - speed

用戶83992

2013-05-08 12:31 pm

A crate of mass M starts from rest at the top of a frictionless ramp inclined at an angle α above the horizontal. Find its speed at the bottom of the ramp, a distance d from where it started. Do this in two ways:

(a) Take the level at which the potential energy is zero to be at the bottom of the ramp with y positive upward. (Use any variable or symbol stated above as necessary.)
v = ____m/s

(b) Take the zero level for potential energy to be at the top of the ramp with y positive upward. (Use any variable or symbol stated above as necessary.)
v = ____m/s

(c) Why did the normal force not enter into your solution?

回答 (1)

天同

2013-05-08 3:44 pm

✔ 最佳答案

(a) Potential energy on top of ramp
= Mg(d.sin(α))
where g is the acceleration due to gravity

Kinetic energy at bottom of ramp
= (1/2)Mv^2

By conservation of energy
(1/2)Mv^2 = Mg(d.sin(α))
hence, v = square-root[2.g.d.sin(α)]

(b) Potential energy on top of ramp = 0

Potential energy at bottom of ramp
= -Mg(d.sin(α))

Kinetic energy at bottom of ramp
= (1/2)Mv^2

By conservation of energy
0 = (1/2)Mv^2 + [-Mg(d.sin(α))]
i.e. (1/2)Mv^2 = Mg.(d.sin(α))
hence, v = square-root[2.g.d.sin(α)]

(c) It is because the normal force doesn't do any work.

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