A block of mass m = 25.0 kg slides on a frictionless inclined track as shown in

Question

A block of mass m = 25.0 kg slides on a frictionless inclined track as shown in the figure. There is a slot cut out along the center of the track so that a massless spring with spring constant k = 5,000. N/m can remain attached to the block at all times. The other end of the spring is at also cylinder that rides on a frictionless, horizontal rail at the bottom of the ramp. The spring is also rigid, so that as the block slides along the ramp, the spring remains perfectly vertical at all times. The rise and run of the incline are both 8.00m. The unstretched, equilibrium position of the spring occurs when the spring is 5.00m from the left end of the incline, as shown in the figure. Let s be the distance of the block along the track from the top end of the ramp. a) If the block is initially pulled up to the very top of the ramp at s = 0, and then let go, what will the speed of the block be when it again gets to the position shown in the figure, which is 5.00m, horizontally, from the left end of the ramp? b) What is the lowest position the block will reach after it is let go? c) Describe what the motion of the block will be after it is let go from the top of the ramp. d) Is there a position along the we can get the block to remain perfectly still? If so, ramp where find what it is.

Explanation / Answer

a) After letting go, when the block reaches at this position (5m horizontally from left end of ramp), the spring is relaxed and so it's potential energy is zero.

Now, let us suppose the height of the block in this position (5m horizontally from left end of ramp) be h.

then h = 5m as the incline's angle is 450 (both rise and run are same.), h is also the natural length of the spring.

So when the block is at s = 0 , the elongation of spring is x = 8m - h = 8m - 5m = 3m

Energy at s = 0 is E1 = Mg(8m) + (1/2)kx2, where x = 3m

Energy at the position (5m horizontally from left end of ramp) = E2 = (1/2)Mv2 + Mg(5m), where v is the speed at this position.

So applying conservation of mechanical energy we have:

Mg(8m) + (1/2)kx2 = (1/2)Mv2 + Mg(5m)

or Mg(3m) + (1/2)(5000N/m)(3m)2 = (1/2)Mv2

or (25kg)(9.8m/s2)(3m) + (1/2)(5000N/m)(3m)2 = (1/2)(25kg)v2

or v = 43.11m/s

***************************************************************************************************

b) Let at the lowest position the block is at a distance x horizontally from left end of ramp. The block's height would also be x. At this position the spring would be compressed and the compression is 5-x.

At this position speed of the block is 0.

Energy at this position for the block is E2 = Mgx + (1/2)k(5-x)2

Energy at s =0 for the block is E2 =  Mg(8m) + (1/2)k(3m)2

So applying conservation of mechanical energy we have:

Mgx + (1/2)k(5-x)2 = Mg(8m) + (1/2)k(3m)2

245x + 2500(5-x)2 = 1960 + 22500

or 245x + 2500(25 + x2 - 10x) = 24460

or 2500x2 - 24755x + 38040 = 0

or x = 8m, 1.902m (we discard the x = 8m value as it is the value at s = 0)

The lowest position the block reaches when it is let go is x = 1.902m horizontally from left end of ramp.

***************************************************************************************************

c) So, as seen above, the motion will oscillate between s=0 and and the position where the block is 1.902m horizontally from left end of ramp.

After reaching x = 1.902m horizontally from left end of ramp, the block will again start its journey up the incline and reach up to s = 0 position and will be again move down the incline from there to reach the position x = 1.902m horizontally from left end of ramp,...... and so on.....

***************************************************************************************************

d) Let us suppose the block is at a position where it is at a distance x horizontally from left end of ramp.

At this position the block remians still, so the force on it should be zero.

Also since the force of gravity on the block is always downwards, at the position, where the block remians still, the spring would have to provide force in upward direction to balance the force of gravity and this will only happen when the spring is compressed.

At a position, where the block is at a distance x horizontally from left end of ramp, the compression in the spring will be 5-x.

So force on the block by the spring is k(5-x).

So for equilibrium Mg = k(5-x)

or 245 = 5000(5-x)

or x = 4.951m

At this position the force of spring and gravity will balance each other. So there is no compnent of unbalanced force along the incline. So the block will not move along the incline. But there would be a normal force on the block perpendiuclar to the incline which can make it go off the incline at this position if the blokc is allowed. This normal force won't cause any movement along the incline.

***************************************************************************************************

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification or correction I will be happy to oblige....

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.