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Dimas [21]
3 years ago
8

An initially stationary 1.3 kg object accelerates horizontally and uniformly to a speed of 9.4 m/s in 3.0 s. (a) In that 3.0 s i

nterval, how much work is done on the object by the force accelerating it
Physics
1 answer:
ehidna [41]3 years ago
7 0

Answer: 57.434 Nm

Explanation: mass of object = 1.3kg, initial speed = 0m/s ( since it was at rest), final speed = 9.4 m/s.

Using the work energy theorem,

The work done on the object equals it kinetic energy.

Work done = mv²/2

Work done = 1.3 ×( 9.4 - 0)²/2

Work done = 1.3 × 9.4² / 2

Work done = 114.868 / 2 = 57.434 Nm

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When friction slows a sliding block___
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A

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3 years ago
A constant force of magnitude F acts on an object of mass 0.04kg initially at rest at a point O. If the speed of the object when
vampirchik [111]

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F = ?

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v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

v^{2} = u^{2} + 2as\\\\=> a = \frac{v^{2} - u^{2}}{2s}\\a = \frac{500^{2} }{2*50}\\a = 2500ms^{-2}

Then F = ma = 0.04 x 2500 = 100N

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3 years ago
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