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Talja [164]
1 year ago
7

If the resultant force acting on a 2.0 kg object is equal to (3.0î + 4.0ĵ) N, what is the change in kinetic energy as the object

moves from (7.0î – 8.0ĵ) m to (11.0î – 5.0ĵ) m?
Physics
1 answer:
12345 [234]1 year ago
3 0

Answer:

ΔK = 24 joules.

Explanation:

ΔK = Work done on the object

Work is equal to the dot product of force supplied and the displacement of the object.

W = F * Δs

Δs can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δs = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).

This gives us

W = < 3, 4 > * < 4, 3 > = (3*4)+(4*3) = 24 J

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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
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Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

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2 years ago
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