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oee [108]
2 years ago
5

A truck of mass 10000kg moving through a sloppy road of length 20km with the help of 5000N force. If the truck climbs the vertic

al height of 200m and it's efficiency is 80% calculate the additional mass that can be carried by the truck​ ?​
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
4 0

Answer:

30000

Explanation:

Statement of the given problem,

A truck of mass 10000 kg is moving through a sloppy road of length 20 km with the help of 5000 N force. If the truck climb the vertical height of 200 m and its efficiency is 80% calculate the additional mass that can be carried by the truck?

Let M denotes the additional mass (in kg) that can be carried by the truck.

From above data we determine that,

P.E. gained by total mass at the vertical height of 200 m = 80% of work done by the applied force [road elevation = tan^-1 (200/20000) = 0.57° (assumed negligible)]

Hence we get following relation,

(10000 + M)*g*200 = 5000*20*1000*80/100 [g = gravitational acceleration]

or 10000 + M = 50000*8/g = 50000*8/10 [g = 10 m/s^2 (assumed)]

or M = 40000 - 10000 = 30000 (kg) [Ans]

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3 years ago
A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha
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Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

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x(t) = A\cdot \cos (\omega \cdot t + \phi)

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x(t) - Position of the mass with respect to the equilibrium position, measured in centimeters.

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\omega - Angular frequency, measured in radians per second.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)

Where the maximum acceleration of the system is represented by \omega^{2}\cdot A.

The natural frequency of the mass-spring system is:

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

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

If k = 12\,\frac{N}{m} and m = 0.40\,kg, the natural frequency is:

\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }

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Lastly, the maximum acceleration of the system is:

a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)

a_{max} = 359.970\,\frac{cm}{s^{2}}

The maximum acceleration of the system is 359.970 centimeters per square second.

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