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On the surface of the earth, the gravitational field (with z as vertical coordinate measured in meters) is F = 0, 0, −g . Beginning at rest, a ball of mass m = 2 kg moves under the influence of gravity (without friction) along a path from P = (3, 2, 400) to Q = (−21, 40, 50). Find the ball's speed when it reaches Q. (Assume g = 9.8 m/s2. Round your answer to one decimal places.)

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lublana

Answer:

82.8 m/s

Explanation:

We are given that gravitational field F=<0,0,-g>

Potential function =V(x,y,z)

F=-[tex]\nabla V[/tex]=gz

Using fundamental theorem for gradient vector field

Work done =[tex]-\int m F\cdot ds[/tex]

Work done=[tex]2(V(-21,40,50)-V(3,2,400))=2\cdot 9.8\times (-350)=-6860 joules[/tex]

By conversion of energy,work done  convert  into kinetic energy

[tex]\frac{mv^2}{2}=6860[/tex]

[tex]v=\sqrt{\frac{2\cdot 6860}{2}}=82.825 m/s[/tex]

Hence, the speed of ball when it reaches=82.8 m/s

johanrusli

The ball's speed when it reaches Q is about 82.8 m/s

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Further explanation

Let's recall the Kinetic Energy formula:

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

Ek = kinetic energy ( J )

m = mass of object ( kg )

v = speed of object ( m/s )

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Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

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Given:

gravitational acceleration = g = 9.8 m/s²

initial height at P = h_p = 400 m

final heght at Q = h_q = 50 m

initial speed at P = v_p = 0 m/s

Asked:

ball's speed at Q = v_q = ?

Solution:

We will calculate the speed of the ball at Q by using Conservation of Energy formula as follows:

[tex]Ep_1 + Ek_1 = Ep_2 + Ek_2[/tex]

[tex]mgh_p + \frac{1}{2}m v_p^2 = mgh_q + \frac{1}{2}m v_q^2[/tex]

[tex]gh_p + \frac{1}{2} v_p^2 = gh_q + \frac{1}{2} v_q^2[/tex]

[tex]2gh_p + v_p^2 = 2gh_q + v_q^2[/tex]

[tex]v_q^2 = 2gh_p - 2gh_q + v_p^2[/tex]

[tex]v_q^2 = 2g( h_p - h_q ) + v_p^2[/tex]

[tex]v_q = \sqrt{ 2g( h_p - h_q ) + v_p^2 }[/tex]

[tex]v_q = \sqrt{ 2(9.8)( 400 - 50 ) + 0^2 }[/tex]

[tex]v_q \approx 82.8 \texttt{ m/s}[/tex]

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Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Energy

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