Question

A box of mass M is released from rest at location A on the track with a trapezoid cross-section as shown below. The elevated region between A and B is smooth and so is the elevated region between C and D. The flat region between B and C is rough; in this region the motion of the box is retarded. The box slides smoothly from A to B with uniform acceleration of magnitude g sin skids from B to C with uniform deceleration of magnitude a_0, and then slides smoothly up from C to rest momentarily at D with acceleration of magnitude g sin theta. At point D, the box reverses its motion, i.e., it begins to move in the opposite direction. Assume the following: the initial location A is at heights h_0 relative to the flat region of the track, AB makes angle theta_0 with the horizontal, CD makes angle theta with the horizontal and the length of the track BC is I. Also assume that the acceleration of the box in each region has a constant magnitude. Answer the questions below. Outline the physical principles employed for each question in a few steps and express all your answers in terms of the given variables (where necessary). (a) What is the total time taken to travel from A to D? (b) What is the speed of the box midway between C and D during the forward motion? (c) What range of values of a0 will guarantee that the box will stop somewhere in the flat region BC during the reverse motion? (d) What is the maximum height, between A and B, reached by the box in the reverse motion (assuming that such a motion is possible)? What is the optimal value that a_0 must attain for this to occur?

Answer 1

a)

let time for A to B t_AB :

AB = d_ab = h_o /Sin $\theta$ _o

a = acceleration = g Sin $\theta$ _o

V_o = initial velocity = 0 m/s

using the equation

d_ab = V_o t_AB + (0.5) a t_AB²

h_o /Sin $\theta$ _o = 0 + (0.5) (g Sin $\theta$ _o) t_AB²

t_AB = sqrt(2h_o /gSin² $\theta$ _o )

V_f = final velocity at B

V²_f = V²_o + 2a d_AB

V²_f = 0 + 2(g Sin $\theta$ _o) (h_o /Sin $\theta$ _o)

V_f = sqrt(2gh_o)

from C to D :

V_o = initial velocity at C = sqrt(2gh_o - 2a_ol)

V_f = final velocity at D = 0

a = - g Sin $\theta$

Using the equation

V_f = V_o + a t

0 = sqrt(2gh_o - 2a_ol) - g Sin $\theta$ t_CD

t_CD = sqrt(2gh_o - 2a_ol) /(g Sin $\theta$ )

for B to C :

V_o = initial velocity = sqrt(2gh_o)

a = - a_o

d = BC = l

using

V_f² = V_o² + 2 a d

V_f² = (sqrt(2gh_o))² - 2 a_o l

V_f = sqrt(2gh_o - 2a_ol)

using the equation

V_f = V_o + at

sqrt(2gh_o - 2a_ol) = sqrt(2gh_o) - a_o t_BC

t_BC = (sqrt(2gh_o) - sqrt(2gh_o - 2a_ol) ) /a_o

t_AD = t_AB + t_BC + t_CD

t_AD = sqrt(2h_o /gSin² $\theta$ _o ) + (sqrt(2gh_o) - sqrt(2gh_o - 2a_ol) ) /a_o    + sqrt(2gh_o - 2a_ol) /(g Sin $\theta$ )

b)

for C to D

Using the equation

V_f = V_o + a t

V_f = sqrt(2gh_o - 2a_ol) - g Sin $\theta$ (sqrt(2gh_o - 2a_ol) /(2g Sin $\theta$ ))

V_f = sqrt(2gh_o - 2a_ol)/2

c)

using

V_f² = V_o² + 2 a d

0² = (sqrt(2gh_o))² - 2 a_o l

a_o = gh_o /l