suppose an inclined plane is tipped at angle = 6.7 o . The 0 cm mark is at the b

Question

suppose an inclined plane is tipped at angle = 6.7 o. The 0 cm mark is at the bottom.

Photogate 1 is at the 27.7 cm mark, and photogate 2 is at the 70.1 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 583 g.
- The track is totally frictionless

Suppose the cart is released at rest at some point above photogate 2 (the higher point on the track). If the cart passes through photogate 2 at speed 0.589 m/s, at what speed will the cart pass through photogate 1?
v1 = ____________ m/s

Explanation / Answer

at photogate 2:

h2 = 70.1sin6.7 = 8.18 cm = 0.0818 m

PE = mgh2 = 0.0818mg

KE = mv^2 /2 = m x 0.589^2 /2 =0.173m

at photogate 1:

h1 = 0.277sin6.7 = 0.0323

PE = mgh1 = 0.0323mg

KE = mv^2 /2

Using energy conservation, PE + KE = constant

0.0818mg + 0.173m = 0.0323mg + mv^2 /2

v = 1.15 m/s .............Ans

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