Acubic shaped box has a side length of 1.0 ft and a mass of 10 lbm is sliding on a frictionless horizontal surface towards a 30 upward incline. the horizontal velocity of the box is 20 ft/s. determine how far up the incline the box will travel (report center of mass distance along the inclined surface, not vertical distance)

Explanation  &

Assuming a smooth transition so that there is no abrupt change in slopes to avoid frictional loss nor toppling, we can use energy considerations.

Initially, the cube has a kinetic energy of

KE = mv^2/2 = 10 lbm * 20^2 ft^2/s^2  / 2 = 2000 lbm-ft^2 / s^2

At the highest point when the block stops, the gain in potential energy is

PE = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2

By assumption, there was no loss in energies, we equate PE = KE

322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2

=>

h = 2000 /322 = 6.211 (ft)

distance up incline = h / sin(30) = 12.4 ft

how is this a question? in order for me to answer it i have to understand the question

explanation:

answer:

dsfdssdfdssfdsfd

explanation: