climb to the top of er coaster travelling towards the right has just suffcient e

Question

climb to the top of er coaster travelling towards the right has just suffcient energs to ekib The heigin of C and E. The height of other losses and assume hill D which is 100 m above the ground level represented by points A hill B above the ground level is 35.714 n Ignore friction or ot conservation of energy. Use g= 9.30 m/sec 110 points, (a) What was the speed of the coaster at A? (b) What was the speed of the coaster at B? (c) At what height above the ground will the coaster have a speed of 10 You MUST work out (a), (b) and (e) in that order. m/s?

Explanation / Answer

While climbing, the kinetic energy of the Roller coaster at the ground converts into potential energy at the top of the Hill D.

Since it has enough energy to reach 100m,

[(1/2) m v2 ] at ground= [m g h ] at hill D

i.e. v at ground= sqrt(g h) at hill D

(a) from above, v at ground, A = sqrt (2x9.80 x 100) = 44.27 m/s

(b) use the third equation of motion, v2 = u2 – 2.g.h

                u = 44.27, g= 9.80, h = 35.714m, v=?

                So v = sqrt (44.272 – 2x9.80x35.714) = 35.5 m/s

(c) again, use the 3rd qequation of motion,

                Here, v= 10m/s, u = 44.27m/s, g=9.80m2/s, h = ?

                102 = 44.272 – 2x9.80xh

                i.e. h = 94.9 m

hence, at about 95m, speed will be 10m/s

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