Physics, 12.10.2019 04:30 Misspaige5150

A2.50-kg mass is pushed against a horizontal spring of force constant 25.0 n/cm on a frictionless air table. the spring is attached to the tabletop, and the mass is not attached to the spring in any way. when the spring has been compressed enough to store 11.5 j of potential energy in it, the mass is suddenly released from rest. find the greatest speed the mass reaches, when does it occur? what is the greatest acceleration of the mass and when does it occur?

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Answer from: chrisraptorofficial

1) Maximum speed: 3.0 m/s, when the spring is at equilibrium position

Explanation:

The law of conservation of energy tells that the total mechanical energy of the system, given by the sum of elastic potential energy (U) and kinetic energy (K), is constant:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2=const.$

where

k is the spring constant

x is the compression/stretch of the spring with respect to its equilibrium position

m is the mass of the block

v is the speed of the block

The maximum speed of the system occurs when the kinetic energy is maximum, and therefore when the elastic potential energy is zero: U=0, so we have:

$U=\frac{1}{2}kx^2=0$

which means when x=0, so when the spring is at its equilibrium position. At this point of the motion, we also have

$E=K_{max}=\frac{1}{2}mv_{max}^2$ (1)

The total mechanical energy of the system is given by the problem:

E=11.5 J

which corresponds to the potential energy of the system when the spring is at its maximum compression (because there the kinetic energy is zero). Solving (1) for v_max, we find:

$v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(11.5 J)}{2.50 kg}}=3.0 m/s$

2) Maximum acceleration: 96 m/s^2 , when the spring is at its equilibrium position

Explanation:

The restoring force on the spring is given by Hook's law:

F=kx

Using Newton's second law, we can rewrite the force as product between mass and acceleration:

ma=kx

so we have

$a=\frac{kx}{m}$

We infer that the greatest acceleration, $a_{max}$ , will occur also when the displacement is maximum, $x_{max}$ :

$a_{max}=\frac{kx_{max}}{m}$

The maximum displacement occurs when the elastic potential energy is maximum and the kinetic energy is zero, K=0, so we have:

$E=\frac{1}{2}kx_{max}^2\\x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(11.5 J)}{2500 N/m}}=0.096 m$

So, the greatest acceleration is

$a=\frac{(2500 N/m)(0.096 m)}{2.5 kg}=96 m/s^2$

Answer from: Quest

Answer: it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.explanation: we have equation of motion $s=ut+\frac{1}{2} at^2$ ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/s $s=0*t+\frac{1}{2} *2.18*t^2$ $s=1.09t^2[/ 2equating equation 1 and 2 [tex]4.75t = 1.09t^2\\ \\ t=4.36 seconds.$ so it will take 4.36 seconds to reach her average velocity 10.69 miles/hour.

Answer from: Quest

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