A cylinder (I = MR²/2)is rolling along the ground at 7.0 m/s.It comes to a hill and starts going up.Assuming no losses to friction,how high does it get before it stops? A)1.2 m B)3.7 m C)4.2 m D)5.9 m

Question

Quizplus · Accepted Answer

The correct answer is B.The cylinder's initial kinetic energy is:$$K_i = \frac{1}{2}MR^2\omega^2 = \frac{1}{2}MR^2\left(\frac{v}{R}ight)^2 = \frac{1}{2}Mv^2$$The cylinder's final potential energy is:$$U_f = Mgh$$Assuming no losses to friction, the initial kinetic energy is equal to the final potential energy:$$\frac{1}{2}Mv^2 = Mgh$$Solving for h, we get:$$h = \frac{v^2}{2g} = \frac{(7.0 	ext{ m/s})^2}{2(9.8 	ext{ m/s}^2)} = 3.7 	ext{ m}$$

A Cylinder (I = MR2/2)is Rolling Along the Ground at 7.0

A Cylinder (I = MR²/2)is Rolling Along the Ground at 7.0