A cart rolls with negligible friction on a ramp that is inclined at an 24 degree

Question

A cart rolls with negligible friction on a ramp that is inclined at an 24 degree angle above level ground. It is released from rest and reaches the bottom of the ramp in 2 seconds. How far did it travel along the ramp? Correct, computer gets: 7.97 m Hint: Recall that acceleration on a tilted ramp is constant and related to Earth's gravitational acceleration by a=gsin() a = g sin ( ) . 21) What was the vertical height from which the cart was released? Correct, computer gets: 3.24 m Hint: This involves some triangle geometry. In this situation the sine of the angle of incline is the height over the hypotenuse. 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 28 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)? Answer: Last Answer: 1.73 m Incorrect, tries 8/20. Hint: You'll need to figure out how fast the cart was going at the bottom of the first ramp in the previous problem. Then, in this scenario note that the acceleration is slowing down the cart rather than speeding it up. 23) What is the maximum vertical height that the cart reaches when it rolls up the second ramp? Answer:

Explanation / Answer

a)

s = u t + 0.5*at^2

So, s = 0 + 0.5*(9.8*sin(24 deg))*2^2

So, s = 7.97 m <-----answer

b)

vertical height = 7.97*sin(24 deg) = 3.24 m

c)

KE at the bottom of the 1st ramp = m*9.8*(3.24)

Let the height reached on the 2nd ramp be h

So, mgh = m*9.8*(3.24) <------ By Energy conservation

So, h = 3.24 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

So, L*sin(28 deg) = 3.24

So, L = 6.9 m <--------answer

d)

maximum vertical height = 3.24 <---- same as in 1st ramp.

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