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A 2.00-kg block is pushed along a horizontal frictionless table a distance of 3.00 m, by a horizontal force of 12.0 N. Find (a) how much work is done by the force, (b) the final kinetic energy of the block, and (c) the final velocity of the block. (d) Using Newton’s second law, find the acceleration and then the final velocity.

Respuesta :

(a) The work is done by the force will be 36 J.

(b) The final kinetic energy of the block will be 36 J

(c) The final velocity of the block will be 6 m/sec.

(d) The acceleration of the block will be 6 m/sec².

What is work done?

Work done is defined as the product of applied force and the distance through which the body is displaced on which the force is applied.

Work may be zero, positive, and negative.it depends on the direction of the body displaced. if the body is displaced in the same direction of the force, it will be positive.

The work done by force is;

[tex]\rm W = Force \times displacement \\\\ W=F \times d \\\\ W= 12 \times 3 \\\\ W= 36 \ joule[/tex]

From work energy theory;

The change in the kinetic energy is equal to the work done by force and the work done by friction;

[tex]\rm W_{force}+W_{friction}=\trinagle KE \\\\ 36+0=KE_{final}-KE_{initial}\\\\ 36+0=KE_{final}-0\\\\ KE_{final}=36\ J[/tex]

The final velocity of the block is found as;

[tex]\rm KE_{FINAL}= \frac{1}{2} mV_{final}^2\\\\ V_{final}= \sqrt{36} \\\\ V_{final}=6 \ m/sec[/tex]

From the Newton'saw;

F=ma

12=2×a

a=6  m/sec²

Hence, the work is done by the force will be 36 J.

To learn more about the work done, refer to the link ;

https://brainly.com/question/3902440

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