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A 1.50-kg book is sliding along a rough horizontal surface. at point a it is moving at 3.21 m>s, and at point b it has slowed to 1.25 m>s. (a) how much work was done on the book between a and b? (b) if -0.750 j of work is done on the book from b to c, how fast is it moving at point c? (c) how fast would it be moving at c if +0.750 j of work was done on it from b to c?

Respuesta :

Given:
Mass, m = 1.50 kg
Velocity at point a, v1 = 3.21 m/s
Velocity at point b, v2 = 1.25 m/s

Part (a)
Because the velocity decreased, work was done to slow down the book. The work is equal to the loss in kinetic energy.
Work done = (1/2)*(1.5 kg)*(3.21² - 1.25² m²/s²) = 6.556 J

Answer: -6.556 j

Part (b)
Let v3 =  velocity of the book at point c.
Work done is -0.75 J from point b to point c.
This will cause loss of kinetic energy of the book.
Therefore
(1/2)*(1.5 kg)*(1.25² - v3² m²/s²) = 0.75 J
0.75(1.5625 - v3²) = 0.75
1.5625  - v3² = 1
v3² = 0.5625
v3 = 0.75 m/s

Answer: The velocity at c is 0.75 m/s

Part (c)
If +0.75 J of work is performed on the book between b and c, the kinetic energy of the book will increase. Therefore
(1/2)*(1.5 kg)*(v3² - 1.25² m²/s²) = 0.75 J
0.75(v3² - 1.5625) = 0.75
v3² - 1.5625 = 1
v3² = 2.5625
v3 = 1.6 m/s

Answer: The velocity at c is 1.6 m/s

johanrusli

(a) -6.56 J of work is done on the book between a and b

(b) If -0.750 J of work is done , then the book is moving as fast as 0.75 m/s

(c) If +0.750 J of work is done , then the book is moving as fast as 1.60 m/s

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Further explanation

Let's recall the formula of Kinetic Energy as follows:

[tex]\large {\boxed {E_k = \frac{1}{2}mv^2 }[/tex]

Ek = Kinetic Energy ( Newton )

m = Object's Mass ( kg )

v = Speed of Object ( m/s )

Let us now tackle the problem !

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Given:

mass of book = 1.50 kg

speed at point a = v_a = 3.21 m/s

speed at point b = v_b = 1.25 m/s

Asked:

(a) work done between point a and b = W_ab = ?

(b) speed of book at point c = v_c = ? → W_bc = -0.750 J

(c) speed of book at point c = v_c = ? → W_bc = +0.750 J

Solution:

Question (a):

[tex]W = \Delta E_k[/tex]

[tex]W_{ab} = E_{kb} - E_{ka}[/tex]

[tex]W_{ab} = \frac{1}{2}m( v_b^2 - v_a^2 )[/tex]

[tex]W_{ab} = \frac{1}{2} \times 1.50 \times ( 1.25^2 - 3.21^2 )[/tex]

[tex]W_{ab} = -6.5562 \texttt{ J}[/tex]

[tex]\boxed {W_{ab} \approx -6.56 \texttt{ J}}[/tex]

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Question (b):

[tex]W = \Delta E_k[/tex]

[tex]W_{bc} = E_{kc} - E_{kb}[/tex]

[tex]W_{bc} = \frac{1}{2}m( v_c^2 - v_b^2 )[/tex]

[tex]-0.750 = \frac{1}{2} \times 1.50 \times ( v_c^2 - 1.25^2 )[/tex]

[tex]-0.750 = 0.75 ( v_c^2 - 1.25^2 )[/tex]

[tex]-0.750 \div 0.75 = v_c^2 - 1.25^2[/tex]

[tex]v_c^2 = -1 + 1.25^2[/tex]

[tex]v_c = \sqrt{0.5625}[/tex]

[tex]\boxed {v_c = 0.75 \texttt{ m/s}}[/tex]

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Question (c):

[tex]W = \Delta E_k[/tex]

[tex]W_{bc} = E_{kc} - E_{kb}[/tex]

[tex]W_{bc} = \frac{1}{2}m( v_c^2 - v_b^2 )[/tex]

[tex]0.750 = \frac{1}{2} \times 1.50 \times ( v_c^2 - 1.25^2 )[/tex]

[tex]0.750 = 0.75 ( v_c^2 - 1.25^2 )[/tex]

[tex]0.750 \div 0.75 = v_c^2 - 1.25^2[/tex]

[tex]v_c^2 = 1 + 1.25^2[/tex]

[tex]v_c = \sqrt{2.5625}[/tex]

[tex]\boxed {v_c \approx 1.60 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

Learn more

  • Impacts of Gravity : https://brainly.com/question/5330244
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  • The Acceleration Due To Gravity : https://brainly.com/question/4189441
  • Newton's Law of Motion: https://brainly.com/question/10431582
  • Example of Newton's Law: https://brainly.com/question/498822

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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