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IrinaVladis [17]
4 years ago
15

A constant horizontal pull acts on a sled on a horizontal frictionless ice pond. the sled starts from rest. when the pull acts o

ver a distance x, the sled acquires a speed v and a kinetic energy k. if the same pull instead acts over twice this distance:

Physics
2 answers:
lozanna [386]4 years ago
7 0
W₁ = F * x = 1/2 * m * v₁²

W₂ = F * 2x = 2 * 1/2 * m * v₂² = 1/2 * m * (√2 * v₂)²

v₁ = √2 * v₂
Papessa [141]4 years ago
3 0

<em>If the same pull instead acts over twice this distance, </em>

C) The sled’s speed will be v√2 and its kinetic energy will be 2K.

\texttt{ }

<h3>Further explanation</h3>

Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.

\large {\boxed {F = ma }

F = Force ( Newton )

m = Object's Mass ( kg )

a = Acceleration ( m )

Let us now tackle the problem !

\texttt{ }

<u><em>Complete Question:</em></u>

<em>A constant horizontal pull acts on a sled on a horizontal frictionless ice pond. The sled starts from rest. When the pull acts over a distance x, the sled acquires a speed v and a kinetic energy K. If the same pull instead acts over twice this distance:</em>

<em>A) The sled’s speed will be 2v and its kinetic energy will be 2K.</em>

<em>B) The sled’s speed will be 2v and its kinetic energy will be K√2.</em>

<em>C) The sled’s speed will be v√2 and its kinetic energy will be 2.</em>

<em>D) The sled's speed will be v√2 and its kinetic energy will be K√2.</em>

<em>E) The sled's speed will be 4v and its kinetic energy will be 2K.</em>

\texttt{ }

<em>We will use following formula to solve this problem:</em>

w = \Delta K

Fx = K - 0

\large {\boxed {K = Fx} }

\texttt{ }

<em>Let's compare the two situations by using above formula:</em>

K_1 : K_2 = F_1x_1 : F_2x_2

K : K_2 = Fx : F(2x)

K : K_2 = 1 : 2

K_2 = 2K

\texttt{ }

K_1 : K_2 = \frac{1}{2}m(v_1)^2: \frac{1}{2}m(v_2)^2

K : 2K = (v)^2: (v_2)^2

1 : 2 = (v)^2: (v_2)^2

(v_2)^2 = 2v^2

v_2 = \sqrt{2v^2}

v_2 = v\sqrt{2}

\texttt{ }

<h3>Conclusion:</h3>

<em>If the same pull instead acts over twice this distance, </em>

<em>C) The sled’s speed will be v√2 and its kinetic energy will be 2K.</em>

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441
  • Newton's Law of Motion: brainly.com/question/10431582
  • Example of Newton's Law: brainly.com/question/498822

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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