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 r
amp. 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?
1 answer:
The image showing the inclined plane and and the angle has been attached.
Answer:
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.
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