A 2.00-kg object has a velocity (6.20 1 - 1.80 ĵ) m/s. (Note: From the definition of the dot product, v² = v · V.) (a) What is its kinetic energy at this moment? 61.88 The kinetic energy depends on the speed, which is the magnitude of the velocity vector. J (b) Find the net work done on the object if its velocity changes to (8.00 î+ 4.00 ĵ) m/s. 50.12 How is the work done by the net force related to the change in kinetic energy? J

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A: a)The required work done is,

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A: (a) Using Work-Energy theorem, net work required is,

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A: Given that,The time : t = 1.5 (sec)The mass of block : m = 28 (kg)The displacement of the block…

Q: It takes you 1.5 s to push a 28 kg box by 4.4 m across a horizontal floor with a constant force of…

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A: W=work done by pushing force=friction work done +increase in KE

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A:

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A:

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A: distance moved (d) = 42.6 meters constant applied force (F) = 118.1 N θ = 37.2ofrictional force (f)…

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A: The work done by the force on the object is

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Q: what is the dot product of C and E, if C = 3.45i - 3.5j + 3.48k and E = -3.6i - 7.11j + 3.13k?

A:

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A: Given, Mass of crocodile, m = 531 kgfinal speed , v = 25.9 km/h =25.9×1000m3600s =7.19 m/sTime taken…

Q: A 2.80-kg object has a velocity (6.20 î - 2.40 ĵ) m/s. (Note: From the definition of the dot…

A:

Q: f his toys of mass 15.0 kg along a rough horizontal surface for a distance of 6.00 m. The tension…

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A: Distance covered by crate = 43.2m Force applied by man =104.3N Force applied by friction = 49N