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A solid sphere of weight 42.0 N rolls up an incline at an angle of 38.0

A solid sphere of weight 42.0 N rolls up an incline at an angle of 38.0

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Answer #1

Given that,

Weight = m g = 42 N

so we get m = 42/9.8 =4.28 kg ; Theta = 30 deg ;V(com) = 7.9 m/s

(a) The KE at the bottom will be the sum of KE(rot) + KE(trans)

KE(rot) = 1/2 I 2 where I = 2/5 m r2 and = 2 x pi x V(com) / r = 2 x pi x 7.9 / r

KE(rot) = 1/2 x 2/5 m r2 x 4 pi2 x (7.9)2 / r2

Ke(rot) = 4/5 x (3.14)2 x 4.28 x (7.9)2 = 2106.9 Joules

The translational KE will be

KE(trans) = 1/2 m v2 = 0.5 x 4.28 x (7.9)2 = 133.6 Joules

KE(total) = 2107 + 133.6 = 2240.6 Joules

(b)Let d be the distance. We know that, at a height h the GPE will be

PE = mgh

we get h =  = 2240 / 46 = 48.69 meters

d = h/sin (theta)

d = 48.69 / sin (38) = 79.1 meters

(c) Yes the answer b depends on the mass of the sphere.

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