Work-Energy Theorem calculations

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The discussion focuses on verifying calculations related to the work-energy theorem for a rock thrown vertically. The user calculates the rock's mass as approximately 2.14 kg and determines its speed just after leaving the ground to be 30.8 m/s. They also seek confirmation on using this initial speed to find the maximum height, which they calculate to be around 48.4 m. The responses confirm that using -21.0 N for the force and 30.8 m/s for the initial velocity is correct, as the work done by the weight opposes the motion. The calculations and methodology are validated as accurate.
Soaring Crane
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Please check to see if my calculations are correct. I am unsure if I used the right values and setup.

You throw a rock of weight 21.0 N vertically into the air from ground level. You observe that when it is a height 14.8 m above the ground, it is traveling at a speed of 25.7 m/s upward.

a. Use the work-energy theorem to find its speed just as it left the ground;Take the free fall acceleration to be g = 9.80 m/s^2 .

m_rock = 21.0 N/ 9.8 m/s^2 = 2.14285 kg

F*d = 0.5*m*(v_2)^2 - 0.5*m*(v_1)^2
14.8 m*(-21 N) - 0.5*m*(v_2)^2 = -0.5*m*(v_1)^2

v_1 = sqrt[(F*d - 0.5*m_rock*(v_2)^2)/(-0.5*m_rock)]
= sqrt[((14.8m*-21.0N) - 0.5*2.14 kg*(25.7 m/s)^2)/(-0.5*2.14 kg)]
= 30.8 m/s ??

b. Use the work-energy theorem to find its maximum height.
Take the free fall acceleration to be g = 9.80 m/s^2 .

F*d = 0 - 0.5*m*(v_1)^2
d = [- 0.5*m*(v_1)^2]/[F] = 0.5*(v_initial)^2/g

For v_initial ,am I supposed to use the value that I found in Part A?

d = 0.5*(30.8 m/s)^2/g = 48.4 m ??

Thanks.
 
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Yes, looks perfectly OK.
 
So it is correct that I used -21.0 N for the force and 30.8 m/s for the initial velocity?

Thanks again.
 
Soaring Crane said:
So it is correct that I used -21.0 N for the force and 30.8 m/s for the initial velocity?

Thanks again.

Yes, it is correct, since the work done by the weight has the opposite direction to the velocity.
 
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