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4 years ago

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A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.89 N

/m. The object is displaced 4.73 m to the right from its equilibrium position and then released, which initiates simple harmonic motion. What is the magnitude of the force acting on the object 3.8 s after it is released? Answer in units of N.

1 answer:

mixas84 [53]4 years ago

4 0

To solve this problem, apply the concepts related to the simple harmonic movement. We will start by finding the angular velocity, essential to find the position of displacement of the spring with the data provided. With displacement we will apply Hooke's law.

A ) Position of the object is

$x = Acos(\omega t)$

The angular frequency of the object is

$\omega = \sqrt{\frac{k}{m}}$

$\omega = \sqrt{\frac{4.89}{1.89}}$

$\omega =1.60851 rad/s$

Replacing at the first equation we have that

$x = (4.89)cos((1.60851)(3.8))$

$x = 4.8188m$

Now the force is

$F = -kx$

$F = - (4.89)(4.8188)$

$F = -23.5639N$

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3 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

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$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

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Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

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$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

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