If the same pull instead acts over twice this distance,
C) The sled’s speed will be v√2 and its kinetic energy will be 2K.
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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.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
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Complete Question:
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:
A) The sled’s speed will be 2v and its kinetic energy will be 2K.
B) The sled’s speed will be 2v and its kinetic energy will be K√2.
C) The sled’s speed will be v√2 and its kinetic energy will be 2.
D) The sled's speed will be v√2 and its kinetic energy will be K√2.
E) The sled's speed will be 4v and its kinetic energy will be 2K.
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We will use following formula to solve this problem:
[tex]w = \Delta K[/tex]
[tex]Fx = K - 0[/tex]
[tex]\large {\boxed {K = Fx} }[/tex]
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Let's compare the two situations by using above formula:
[tex]K_1 : K_2 = F_1x_1 : F_2x_2[/tex]
[tex]K : K_2 = Fx : F(2x)[/tex]
[tex]K : K_2 = 1 : 2[/tex]
[tex]K_2 = 2K[/tex]
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[tex]K_1 : K_2 = \frac{1}{2}m(v_1)^2: \frac{1}{2}m(v_2)^2[/tex]
[tex]K : 2K = (v)^2: (v_2)^2[/tex]
[tex]1 : 2 = (v)^2: (v_2)^2[/tex]
[tex](v_2)^2 = 2v^2[/tex]
[tex]v_2 = \sqrt{2v^2}[/tex]
[tex]v_2 = v\sqrt{2}[/tex]
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If the same pull instead acts over twice this distance,
C) The sled’s speed will be v√2 and its kinetic energy will be 2K.
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Grade: High School
Subject: Physics
Chapter: Dynamics
