Arrow energy transformation and two energy/power problems

[jakeowens]

An arrow is fired, via a bow, straight up. It rises for a while and then drops back to the ground. The process, taking the arrow from loading to just prior to touch down, can best be described by a series of energy transformations corresponding to which of the following?
none of these
KE > gravitational PE > work
elastic PE > gravitational PE > KE
work > KE > elastic PE > KE
* work > elastic PE > KE > gravitational PE > KE *

For this one, the correct answer is elastic work > elastic PE > KE > gravitational PE > KE right?

Because you have to do work to pull the string back, then there is elastic potential energy stored in the string, then when it is released it turns into KE, then at the top it turns into gravitational PE, and as it is faling down it the energy has turned back into KE correct?

While traveling along at 96 km/h, a 14.4 kN auto runs out of gas 16 km from a service station. Neglecting friction, if the station is on a level 15.9 m above the elevation where the car stalled, how fast will the car be going when it rolls into the station, if in fact it gets there?

I really don't know exactly where to go with this question. I used the formula KE=1/2mv^2 to figure out the kinetic energy of the car when it runs out of gas, and got 521,654.9N, but I don't know where to go with it from there. The angle of the road is 0.5694 degrees where the car is at, but I'm having trouble figuring out where to go next with this problem. work energy theory maybe? I think i know how to do this problem, its just that the fact that the station is 15.9m higher than the car really throws me off. I don't know how to take that into account.

A 2-hp motor drives a hoist that can raise a load of 30 kg to a height of 25 m. At full power, how long will the hoist take to do it?

This question has me completely baffled. I have no idea what the equivalent power in J or N that 2hp would be, so i have no idea how to even get started. Anyone know what 2hp is equivalent to in some units that i can actually use?

Thanks

 

[Doc Al]

An arrow is fired, via a bow, straight up. It rises for a while and then drops back to the ground. The process, taking the arrow from loading to just prior to touch down, can best be described by a series of energy transformations corresponding to which of the following?
none of these
KE > gravitational PE > work
elastic PE > gravitational PE > KE
work > KE > elastic PE > KE
* work > elastic PE > KE > gravitational PE > KE *

For this one, the correct answer is elastic work > elastic PE > KE > gravitational PE > KE right?

Because you have to do work to pull the string back, then there is elastic potential energy stored in the string, then when it is released it turns into KE, then at the top it turns into gravitational PE, and as it is faling down it the energy has turned back into KE correct?

Sounds good to me.

While traveling along at 96 km/h, a 14.4 kN auto runs out of gas 16 km from a service station. Neglecting friction, if the station is on a level 15.9 m above the elevation where the car stalled, how fast will the car be going when it rolls into the station, if in fact it gets there?

I really don't know exactly where to go with this question. I used the formula KE=1/2mv^2 to figure out the kinetic energy of the car when it runs out of gas, and got 521,654.9N, but I don't know where to go with it from there. The angle of the road is 0.5694 degrees where the car is at, but I'm having trouble figuring out where to go next with this problem. work energy theory maybe? I think i know how to do this problem, its just that the fact that the station is 15.9m higher than the car really throws me off. I don't know how to take that into account.

Consider changes in the gravitational PE of the car.

A 2-hp motor drives a hoist that can raise a load of 30 kg to a height of 25 m. At full power, how long will the hoist take to do it?

This question has me completely baffled. I have no idea what the equivalent power in J or N that 2hp would be, so i have no idea how to even get started. Anyone know what 2hp is equivalent to in some units that i can actually use?

Look up how to convert hp (horsepower) into standard units for power, which is the Watt = J/s -- not Joules (energy) or Newtons (force).

 

[jakeowens]

but isn't the gravitational PE for an object that is suspended in air? not just on a hill?

Or can i just take the change in PEgrav=228,961.57N, and subtract that from the KE of the car which would give me 292,693.38N and use that to figure out the speed of the car from KE=1/2mv^2? which gives me 19.96m/s and converting that back to km/h i get 71.89km/h

If i run it all through like that i get that the car would be going 71.89km/h which seems way to high. Or is that right because friction is being ignored?

For the last problem i got that 1hp=746w, so the engine can produce 1492w.

so 1492w=(294.3N(30kg object)*25m)/t

and solving for t gives me 4.93 seconds. and the units all check out so i believe this one is correct.

Thanks

 

[Doc Al]

but isn't the gravitational PE for an object that is suspended in air? not just on a hill?

The gravitational PE only depends on the change in height of the object. It doesn't matter whether it's in the air, on a hill, or buried underground.

Or can i just take the change in PEgrav=228,961.57N, and subtract that from the KE of the car which would give me 292,693.38N and use that to figure out the speed of the car from KE=1/2mv^2? which gives me 19.96m/s and converting that back to km/h i get 71.89km/h

If i run it all through like that i get that the car would be going 71.89km/h which seems way to high. Or is that right because friction is being ignored?

You got it.

For the last problem i got that 1hp=746w, so the engine can produce 1492w.

so 1492w=(294.3N(30kg object)*25m)/t

and solving for t gives me 4.93 seconds. and the units all check out so i believe this one is correct.

Sounds good to me.