Question

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 7.8 m/s. Ignore frictional losses. (a) What is the height of the hill (6) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom? (a) Number Units (b) Number Units Click if you would like to show Work for this question: Open Show Work

Answer 1

Dear student,

Find this solution, and RATE IT ,If you find it is helpful .your rating is very important to me.If any incorrectness ,kindly let me know I will rectify them soon.

Thanks for asking ..
KIE 5.1734m 9.8 (a) using wook-energy theorem, (Assuming basketball a hollow sphere". I migh = 1 h ² + 2 x 2 m 2 x (x 2 = 5v2 Translational Rotational share=&7-8)2 – 5.1734m And 10% for frozen Juice (taken as solid Cylinder) I = MR² again Hugh = 2 kV? + b (he) (y) ² = 32 4x9,8X501734 ve fish = 4*3-2X51234 v=8.222 m/sec) And (2) h=0.76m v= 6.13 m/sec + xhx² + x 18 8 x) = nhang + 1 h) (4) figh => 7x(613)14? + 9.8x0,76 - V = 5.19 mhree And

③ For solid sphere Wi=21 rad, Torone due to friction (2) Brak I = 2 mR? -0.450 N-m. = 2x1.5x(0.44)² = 0.11616 L Retardation Angular (2) kg-m2 = 2 2 2 = -3.874 rad seca . Wg = Witat 0=21-3.874xt t = time = 5.421 sec For Hollow Sphere, I = 2 x 1.5x10-44) ² - 2 = -0.45 = -2.324388ad 0.1936 see? W = 0=21-2.32438xt t=9.0347 Sec