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denpristay [2]
3 years ago
15

You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 77.2 kg hop on board for a ride

through the woods and the springs (one for each wheel) each compress by 6.37 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 2.14 s. Determine the following.
(a) force constant of the springs 2274.43 X How is the force constant of a spring related to the force the spring exerts when it is stretched or compressed? N/m
(b) mass of the trailer kg
(c) frequency of the oscillation Hz
(d) time it takes for the trailer to bounce up and down 10 times.
Physics
1 answer:
lions [1.4K]3 years ago
5 0

To solve this problem it is necessary to apply the kinematic equations of motion and Hook's law.

By Hook's law we know that force is defined as,

F= kx

Where,

k = spring constant

x = Displacement change

PART A) For the case of the spring constant we can use the above equation and clear k so that

k= \frac{F}{x}

k = \frac{mg}{x}

k= \frac{77.2*9.8}{0.0637}

k = 11876.92N/m

Therefore the spring constant for each one is 11876.92/2 = 5933.46N/m

PART B) In the case of speed we can obtain it through the period, which is given by

T = \frac{2\pi}{\omega}

Re-arrange to find \omega,

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{2.14}

\omega = 2.93rad/s

Then through angular kinematic equations where angular velocity is given as a function of mass and spring constant we have to

\omega^2 = \frac{k}{m}

m = \frac{k}{\omega^2}

m = \frac{ 11876.92}{2.93}

m = 4093.55Kg

Therefore the mass of the trailer is 4093.55Kg

PART C) The frequency by definition is inversely to the period therefore

f = \frac{1}{T}

f = \frac{1}{2.14}

f = 0.4672 Hz

Therefore the frequency of the oscillation is 0.4672 Hz

PART D) The time it takes to make the route 10 times would be 10 times the period, that is

t_T = 10*T

t_T = 10 *2.14s

t_T = 21.4s

Therefore the total time it takes for the trailer to bounce up and down 10 times is 21.4s

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