A1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x axis is applied to the block. the force is given by f with arrow(x) = (2.4 − x2)i hat n, where x is in meters and the initial position of the block is x = 0.
(a) what is the kinetic energy of the block as it passes through x = 2.0 m?
(b) what is the maximum kinetic energy of the block between x = 0 and x = 2.0 m?
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Ответ:
Answer with Explanation:
Mass of block=1.1 kg
Th force applied on block is given by
F(x)=![(2.4-x^2)\hat{i}N](/tpl/images/0387/7710/51a2f.png)
Initial position of the block=x=0
Initial velocity of block=![v_i=0](/tpl/images/0387/7710/5216a.png)
a.We have to find the kinetic energy of the block when it passes through x=2.0 m.
Initial kinetic energy=![K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0](/tpl/images/0387/7710/35e4e.png)
Work energy theorem:
Where
Final kinetic energy
Substitute the values then we get
Because work done=![Force\times displacement](/tpl/images/0387/7710/4175d.png)
Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J
b.Kinetic energy =![K=2.4x-\frac{x^3}{3}](/tpl/images/0387/7710/a0387.png)
When the kinetic energy is maximum then![\frac{dK}{dx}=0](/tpl/images/0387/7710/740e8.png)
Substitute x=![\sqrt{2.4}](/tpl/images/0387/7710/f31c0.png)
Substitute x=![-\sqrt{2.4}](/tpl/images/0387/7710/458e3.png)
Hence, the kinetic energy is maximum at x=![\sqrt{2.4}](/tpl/images/0387/7710/f31c0.png)
Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by
Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J
Ответ:
a
Explanation: