20) A cart rolls with negligible friction on a ramp that is inclined at a 19 deg

Question

20) A cart rolls with negligible friction on a ramp that is inclined at a 19 degree angle above level ground. It is released from rest and reaches the bottom of the ramp in 3 seconds. How far did it travel along the ramp?

21) What was the vertical height from which the cart was released?

22) At the bottom of the first ramp, the cart smoothly rolls onto a second ramp without losing any significant amount of speed. The second ramp is tilted up above the ground by an angle of 24 degrees. What distance up the ramp does the cart reach before coming to rest (give the distance as measured from the base of the second ramp)?

23) What is the maximum vertical height that the cart reaches when it rolls up the second ramp?

Explanation / Answer

20).

given

theta = 19 degree

t = 3 sec

using equation

S = u t + 1/2 a t^2

and we have a = g sin(theta)

S = u t + 1/2 (g sin(theta)) t^2

S = 0 * 3 + 0.5 * ( 9.8 * sin(19) ) * 3^2

S = 0 + 14.3575

S = 14.3575 m or 14.36 m

21)

vertical height = 14.36*sin(19) = 4.68 m

22)

KE at the bottom of the 1st ramp = m*9.8*(4.68)

Let the height reached on the 2nd ramp be h

So, mgh = m*9.8*(4.68) <------ By Energy conservation

So, h = 4.68 m <----- It will go the same height up the ramp

Now, let the length of the ramp upto which it goes up the ramp be L

L*sin(24 deg) = 4.68

L*0.407 = 4.68

L = 11.5 m

23)

maximum vertical height = 4.68 <---- same as in 1st ramp.

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