A block is placed on a frictionless ramp at a height of 11.5 m above the ground. Starting from rest, the block slides down the ramp. At the bottom of the ramp, the block slides onto a frictionless horizontal track without slowing down. At the end of the horizontal track, the block slides smoothly onto a second frictionless ramp. How far along the second ramp does the block travel before coming to a momentary stop, as measured along the incline of the ramp?After the block comes to a complete stop on the second ramp, it will then begin moving back down the second ramp. What is the speed of the block when it is 6.25 m, vertically, above the ground?

The image showing the inclined plane and and the angle has been attached.

The speed of the block when it is 6.25 m, vertically, above the ground is 10.15 m/s

Explanation:

A) Distance the block travels before coming to a momentary stop;

The block will rise to the same height it started at. Hence, the distance along second ramp is;

D = 11.5/sin21.5° = 11.5/0.3665 = 31.38m

B) From the question the ramp is frictionless, and thus we can use conservation of potential and kinetic energy to determine the velocity at the bottom of the first ramp.

In conservation of energy,

K1 + U1 = K2 + U2

Where K1 = initial kinetic energy,

K2 = Final kinetic energy

U1 = Initial potential energy

U2 = Final potential Energy

At this first ramp, initial kinetic energy and final potential energy are both zero, thus U1 = K2

Initial Potential energy(U1) = Mgh

So, U1 = M x 9.81 x 11.5 = 112.82M

Final Kinetic Energy(K2) = ½Mv²

So, from equation of conservation of energy,

112.82M = ½Mv²

Now, M will cancel out,so,

112.82 = ½v²

Multiply both sides by 2 to get ;

225.64 = v²

Thus,

v = √225. 64

v = 15.02 m/s.

Since the, second ramp is also frictionless , we can use the equation of conservation of potential and kinetic energy used earlier to determine the velocity at a height of 6.25 m.

Thus,

Final Potential Energy(U2) = M x 9.81 x 6.25 = 61.81M

Now, for us to determine the block’s kinetic energy at this height position, we have to determine the change in Potential energy from the beginning on the first ramp and now on the second ramp

Thus,

KE = 112.82M – 61.81M = 51.51M

½Mv² = 51.51M

M will cancel out and thus,

½v² = 51.51

Multiply both sides by 2 to get;

v² = 51.51 x 2 = 103.02

v = √103.02 = 10.15 m/s

This is approximately 10.15 m/s.

im guessing the west of station b