Suppose the inclined plane is tipped so it makes an angle = 11 o with the horizo

Question

Suppose the inclined plane is tipped so it makes an angle = 11o with the horizontal. The 0 cm mark is at the bottom. The lower photogate is at the 22.9 cm mark, and the upper photogate is at the 68.7 cm mark along the track, Assume
- The bottom of the track is at U = 0 J (potential energy is zero).
- The total mass of the cart is 147 g.
- The track is totally frictionless

NOTE: Be careful of units!
a) Find the potential energy of the cart
- At the lower photogate: U1 = ___ J
- At the upper photogate: U2 = ___ J

b) Suppose the cart is released at rest at some point above the upper photogate. If the cart passes through the upper photogate at speed 0.567 m/s, at what speed will the cart pass through lower photogate?
v1 = ___ m/s

Explanation / Answer

given,

angle = 11 degree

lower photogate = 22.9 cm

upper photogate = 68.7 cm

mass = 147 g

potential energy at lower photogate = mgh

potential energy at lower photogate = 0.147 * 9.8 * 0.229 * sin(11)

potential energy at lower photogate = 0.0629 J

potential energy at upper photogate = 0.147 * 9.8 * 0.687 * sin(11)

potential energy at upper photogate = 0.1888 J

by conservation of energy

initial energy = final energy

0.147 * 9.8 * 0.687 * sin(11) + 0.5 * 0.147 * 0.567^2 = 0.147 * 9.8 * 0.229 * sin(11) + 0.5 * 0.147 * v^2

v = 1.4263 m/s

cart will pass through lower photogate with speed = 1.4263 m/s

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