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3 years ago

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If a 20 N force acts on a 10 kg object for 0.5 seconds. If the object starts at rest, what would it’s new p (momentum) and v (ve

locity) be?

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

$p=10\:\text{kgm/s}},\\v_f=1\:\text{m/s}$

Explanation:

From Newton's 2nd Law, we have $\Sigma F=ma$ . We can use this to find the acceleration of object after 20 N (force) is applied to the 10 kg object.

Substituting given values, we have:

$\Sigma F=ma, \\20=10a,\\a=\frac{20}{10}=2\:\mathrm{m/s^2}$

Now that we have acceleration, we can find the final velocity of object (after 0.5 seconds) using the following kinematics equation:

$v_f=v_i+at$ , where $v_f$ is final velocity, $v_i$ is initial velocity, $a$ is acceleration, and $t$ is time.

Solving for final velocity:

$v_f=0+2\cdot 0.5,\\v_f=\boxed{1\:\text{m/s}}$

The momentum of an object is given as $p=mv$ . Since we've found the final velocity and mass stays constant, we have:

$p=10\cdot 1=\boxed{10\:\text{kgm/s}}$

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1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

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k = 7 N/m is the spring constant

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Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

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We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

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We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

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Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

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