Simple Energy Problem Im missing something

  • Thread starter Thread starter Stevedye56
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
The discussion centers on calculating the force exerted by the ground on a 20.0 kg coyote that falls 30.0 m and creates a dent of 0.420 m upon impact. The initial potential energy is calculated as 6,000 J, which is converted into kinetic energy just before hitting the ground. Participants clarify that the force exerted by the ground is not simply the weight of the coyote (mg) but involves the work done during the impact. The key point is to consider both gravitational force and the opposing force from the ground to determine the net force during the collision. Understanding this interaction is crucial for accurately calculating the force involved.
Stevedye56
Messages
402
Reaction score
0
The coyote, whose mass is 20.0kg, is chasing the roadrunner when the coyote accidentally runs off the edge of a cliff and plummets to the ground 30.0m below. What force does the ground exert on the coyote as he makes a coyote-shaped dent .420m deep in the ground?

I know that he hits the ground with a force of 6,000J becasue the PE=mgh at the top, Assuming that none of his energy is lost to air resistance he will hit the ground with that same force because all of the potential energy will be converted into kinetic energy. I am just not sure how to get the force. I know F=ma but that seems just to simple.


Any help at all will be appreciated.

Thanks for looking,

Steve
 
Physics news on Phys.org
Stevedye56 said:
What force does the ground exert on the coyote as he makes a coyote-shaped dent .420m deep in the ground?

Do you mean, what is the force which the ground exerts on the coyote while he's been slammed into the ground? If so, use the fact that the change of kinetic energy from the position above the ground to the position when the coyote is stuck in the ground (where his speed = 0) equals the sum of work of gravity (i.e. the coyote's weight) and the work of the force which the ground exerts on him, along the displacement of 0.42 [m]. I hope I didn't miss something.
 
radou said:
Do you mean, what is the force which the ground exerts on the coyote while he's been slammed into the ground? If so, use the fact that the change of kinetic energy from the position above the ground to the position when the coyote is stuck in the ground (where his speed = 0) equals the sum of work of gravity (i.e. the coyote's weight) and the work of the force which the ground exerts on him, along the displacement of 0.42 [m]. I hope I didn't miss something.


Im not sure i understand what your are saying. Are you saying that the force is just mg? Or am i not reading correctly
 
Stevedye56 said:
Im not sure i understand what your are saying. Are you saying that the force is just mg? Or am i not reading correctly

No, the force is not just mg. While being slammed into the ground, the forces acting on the coyote are gravity, and the force which the ground exerts on the coyote (in the opposite direction than gravity), which is the force you need to find.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top