The kinetic energy of the body in definitive position is 4.24 J.
Explanation:
As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.
Frictional force = Coeffficent of friction × Normal force.
As the weight of the body is 2 kg, the normal force acting on it will be mass multiplied with acceleration due to gravity.
Frictional force = - 0.8×9.8 × 2 =-15.68 N
So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.
Work done = Frictional force × Displacement
Work done = -15.68×0.75 = -11.76 J.
So the work is done by the object.
If the kinetic energy of the body at starting is 16 J, then the kinetic energy of the body at definitive position will be obtained as below.
Work done = change in kinetic energy
-11.76 J = Final kinetic energy-16 J
Final Kinetic energy = - 11.76+16
Final kinetic energy = 4.24 J
Thus, the kinetic energy of the body in definitive position is 4.24 J.
: is the initial velocity of the<em> lab cart </em>
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: is the final velocity of the<em> lab cart </em>
: is the <em>final velocity</em> of the <em>object </em>