Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

The drawing shows two frictionless inclines that begin at ground level (h = 0 m)

ID: 1691233 • Letter: T

Question

The drawing shows two frictionless inclines that begin at ground level (h = 0 m) and slope upward at the same angle ?. One track is longer than the other, however. Identical blocks are projected up each track with the same initial speed v0. On the longer track the block slides upward until it reaches a maximum height H above the ground. On the shorter track the block slides upward, flies off the end of the track at a height H1 above the ground, and then follows the familiar parabolic trajectory of projectile motion. At the highest point of this trajectory, the block is a height H2 above the end of the track. The initial total mechanical energy of each block is the same and is all kinetic energy. The initial speed of each block is v0 = 7.77 m/s, and each incline slopes upward at an angle of ? = 50.0°. The block on the shorter track leaves the track at a height of H1 = 1.25 m above the ground. Find (a) the height H for the block on the longer track and (b) the total height for the block on the shorter track.



I got 3.08 m for (a) but no matter how many ways I try to do (b) it comes out wrong! I know that it is supposed to be 1.25 + H2 and that the total is supposed to be less than 3.08. Please show steps with numbers!!!

Explanation / Answer

The block begins with all kinetic energy and leaves the ramp with both potential and kinetic. We can use this knowledge to find the velocity of the block as it leaves the ramp. (1/2)mv^2 = mgh + (1/2)mv^2 (1/2)v^2 - gh = (1/2)v^2 v = sqrt( v^2 - 2gh ) = sqrt( (7.77)^2 - 2(9.8)(1.25) ) = 5.99 m/s Since we are interested in the height obtained at the top of the projectile's path, we solve the following equation for when the vertical component of the velocity is zero. v^2 = v^2 + 2ay y = ( v^2 - v^2 )/(2a) = ((0)-(5.99sin(50))) / (2(-9.8)) = 0.23m For a total height of 1.25+0.23 = 1.48m

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote