1. A mass m is attached to a spring which is held stretched a distance x max by

Question

1. A mass m is attached to a spring which is held stretched a distance xmax by a force Fmax (see figure), and then released. The spring compresses, pulling the mass. For the following answers, assume that there is no friction. Use m for the mass, x for the distance xmax and F for the force Fmax.

a) Determine the speed of the mass m when the spring returns to its normal length (x = 0).

b) Determine the speed of the mass m when the spring returns to half its original extension (x = xmax/2).

2. A 3.0 m long steel chain is stretched out along the top level of a horizontal scaffold at a construction site, in such a way that x = 1.8 m of the chain remains on the top level and y = 1.2m hangs vertically. (See the figure.) At this point, the force on the hanging segment is sufficient to pull the entire chain over the edge. Once the chain is moving, the kinetic friction is so small that it can be neglected. How much work is performed on the chain by the force of gravity as the chain falls from the point where 1.8 m remains on the scaffold to the point where the entire chain has left the scaffold? (Assume that the chain has a linear weight density of 29 N/m.)

Explanation / Answer

a)the potential energy before release is: (1/2)kx2,

at normal length, PE = 0, as there is no extension

let the speed is v, then the kinetic energy = Elastic potential energy (EPE)

i.e (1/2)mv2 = (1/2)kx2

so velcoity V = (kx2/m) = [(k/m)]x

b) similarly agian

Now let the speed is u, from Energy conservation,

we have: (1/2)kx2 = (1/2)k(x/2)2 +(1/2)mu2

u = {[kx2 - kx2/4]/m}

u = [(3k/m)]*(x/2)

-----------------------------------------------

Work done gy gravity = loss in PE

Initila PE = (m/2) g h./2 ) top + (m/2 * g h/2) side

W = 3 mgy/4 *1.5 = 4.5 *29*3 /4 = 97.87 N

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

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