Work-Energy Theorem calculations

In summary, the conversation involves a person asking for help with their calculations regarding a rock being thrown vertically into the air. They use the work-energy theorem to find the rock's speed and maximum height, and confirm their use of force and initial velocity values. The expert confirms that their calculations are correct and explains the reasoning behind them.
  • #1
Soaring Crane
469
0
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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  • #2
Yes, looks perfectly OK.
 
  • #3
So it is correct that I used -21.0 N for the force and 30.8 m/s for the initial velocity?

Thanks again.
 
  • #4
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.
 

FAQ: Work-Energy Theorem calculations

What is the Work-Energy Theorem?

The Work-Energy Theorem is a fundamental principle in physics that states that the work done on an object is equal to the change in its kinetic energy. It is represented by the equation W = ΔKE, where W is work, and ΔKE is the change in kinetic energy.

How is the Work-Energy Theorem calculated?

To calculate the work done on an object, you need to multiply the force applied to the object by the displacement in the direction of the force. This can be represented by the equation W = F * d. To calculate the change in kinetic energy, you need to subtract the initial kinetic energy from the final kinetic energy. This can be represented by the equation ΔKE = KEfinal - KEinitial. Then, you can use the Work-Energy Theorem equation, W = ΔKE, to find the work done on the object.

What are the units of measurement for the Work-Energy Theorem?

The unit for work is joules (J), and the unit for kinetic energy is also joules. So the unit for the Work-Energy Theorem is also joules. However, the units can vary depending on the system of units used (e.g. metric or imperial).

Can the Work-Energy Theorem be applied to all types of motion?

Yes, the Work-Energy Theorem can be applied to all types of motion, including linear, rotational, and oscillatory motion. It is a universal principle that helps to understand the relationship between work and energy in various physical systems.

What are some real-life examples of the Work-Energy Theorem in action?

Some common examples of the Work-Energy Theorem in action include pushing a cart, throwing a ball, and using a pendulum. In each of these scenarios, work is done on the object, and its kinetic energy changes, demonstrating the principle of the Work-Energy Theorem.

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