A block starts at rest and slides down a friction-less track expectfor a small r

Question

A block starts at rest and slides down a friction-less track expectfor a small rough area on a horizontal section of the track (asshown in the figure below).
http://i41.tinypic.com/1zozqps.jpg
It leaves the track horizontally, flies through the air, andsubsequently strikes the ground. The acceleration of gravity is9.8m/s2.

A) What is the speed 'v' of the block as it leaves the track?Answer in units of m/s?

B) What is the horizontal distance 'x' the block travels throughthe air? Answer in units of m.

C) What is the total speed of the block as it hits the ground?Answer in units of m/s.

Explanation / Answer

A) We know that energy is conserved, or E = 0. Since E = P + K, it follows that P +K = 0, or K = -P. The change in potentialenergy is equal to mgh = (0.427 kg)(9.8 m/s2)(1.8m- 4.9m) = -13.0J. Therefore, K = -(-13.0J) =13.0J. The block then runs along a section of length 1.1m with acoefficient of friction equal to 0.4. The force of frictionis equal to N. Since the block is not acceleratingvertically, we know that in the vertical direction, F = N -mg = 0, so it follos that N = mg. Therefore, the frictionalforce is equal to mg. The work done by the frictionalforce is then equal to mgd, which is equal to (0.4)(0.427kg)(9.8 m/s2)(1.1 m) = 1.84 J. This is opposite ofthe direction of motion, so the kinetic energy of the block afterit passes through that section is equal to 13.0J - 1.84J =11.1J. Now we have that: K = mv2/2 = 11.1J v2 = 22.3J/m = 22.3J/(0.427 kg) = 52.1m2/s2 v = 7.2 m/s v = 7.2 m/s B) First, we must determine the length of time that the blockspends in the air. This will be done by usingkinematics. The block has to fall 1.8m more. We use theequation x = vot + at2/2. Sincethe initial vertical velocity is zero, this simplifies to x =at2/2, or: t2 = 2x/a = 2(1.8 m)/(9.8 m/s2)= 0.37 s2 t = 0.61 s From there, to find the horizontal distance traveled, sincethere is no acceleration in the horizontal direction, we use: d = vt = (7.2 m/s)(0.61 s) = 4.4 m C) When the block hits the ground, it will have converted therest of its potential energy to kinetic energy. When it justleaves the ramp, the block has a kinetic energy of 11.1J, and apotential energy equal to mgh = (0.427 kg)(9.8 m/s2)(1.8m) = 7.5J. Thus, when the block hits the ground, it haskinetic energy of 11.1J + 7.5J = 18.7J. Now, we solve forfinal velocity by using: K = mv2/2 = 18.7J v2 = 37.3J/(0.427 kg) = 87.4m2/s2 v = 9.3 m/s v = 9.3 m/s

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