1. In the figure, a spring with a spring constant k is at the top of a frictionl

Question

1. In the figure, a spring with a spring constant k is at the top of a frictionless incline of angle ?. The lower end of the incline is a distance D from the end of the spring, which is at its relaxed length. An object with mass m is pushed against the spring until the spring is compress a distance x. The object is then released from rest.
When the object loses contact with the spring, it has a speed v2. When the object reaches the bottom of the incline, it has a speed v3. Set the gravitational potential energy equal to zero at the bottom of the incline. The goal of this problem is to determine v2 and v3.
a. Determine the kinetic energy of the object, gravitational potential energy, and elastic potential energy at each of the following locations:
i. Where the object is released.
ii. Where the object loses contact with the spring.
iii. At the bottom of the incline plane.
Put all answers in terms of m, k, D, x, ?, v2, v3, and physical constants. (4 points)
b. Using your answers from part a, write an expression for the total mechanical energy at each of the three locations listed in part a:
i. Where the object is released.
ii. Where the object loses contact with the spring.
iii. At the bottom of the incline plane
Put answer in terms of m, k, D, x, ?, v2, v3, and physical constants. (2 points)
c. Using conservation of energy and your answers from parts a and b, determine the speed v2 of the object when it loses contact with the spring. Put answer in terms of m, k, x, ?, and physical constants. (2 points)
d. Using conservation of energy and your answers from parts, a, and b, determine the speed v3 of the object when it reaches the bottom of the incline. Put answer in terms of m, k, D, x, ?, and physical constants. (2 points)

PLEASE SHOW WORK!!!

Explanation / Answer

When it is compressed by a distance x

It is at a distance of D+x from the bottom of the incline

means at a height of (D+x)sin(theta)

i)

KE at this point = ZERO

g = acceleration due to gravity

PE = mgh = mg*(D+x)sin(theta)

Elastic Potential Energy EPE = 0.5kx^2

ii)

When object looses contact it is at a height of Dsin(theta)

KE = 0.5 * mv2^2

PE = mgh = mgDsin(theta)

EPE = zero

iii)

When it is at the bottom spring compression is zero ,

KE = 0.5*mv3^2

PE = ZERO

EPE = ZERO

2)

i)

Total Mechanical Energy(TME1) =KE + PE + EPE = mg*(D+x)sin(theta) + 0.5kx^2

ii)

Total Mechanical Energy(TME2) =KE + PE + EPE = 0.5 * mv2^2+ mgDsin(theta)

iii)

Total Mechanical Energy(TME3) =KE + PE + EPE= 0.5 * mv3^2

3)

using conservation of energy

TME1 =TME2 = TME3

mg*(D+x)sin(theta) + 0.5kx^2 = 0.5 * mv2^2+ mgDsin(theta)

mgx*sin(theta) + .5 kx^2 = .5 mv2^2

v2 = sqrt( 2gx*sin(theta)   + kx^2 / m)

4)

mg*(D+x)sin(theta) + 0.5kx^2 = 0.5 * mv3^2

v3 = sqrt( 2g(D+x)*sin(theta) + kx^2 /m)

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