PHYSICS - ENERGY STORES QUESTON

[sahra]

idk if my answer is right, could someone please be kind enough to tell me if its right?

the question is A 0.1 KG TOY CONTAINS A COMPRESSED SPRING. WHEN THE SPRING IS RELEASES, THE TOY FLIES 0.45M UPWARDS FROM GROUND LEVEL, BEFORE FALLING BACK DOWN TO THE GROUND. ASSUMING THERE IS NO AIR RESISITANCE, CALCULATE THE SPEED OF THE TOY WHEN IT HITS THE GROUND. GRAVATIONAL FIELD STRENGH IS 9.8 N/KG. here's my working out...

1) i figured out the gravational potential energy = to the kinetic energy in this question

2)gravational potential energy formula is  = mgh so i did o.1 times 0.45 times 9.8 which gave me the answer 0.44j

3) i now figured  i need to work out the speed of the toy meaning i need to re- arrange the kinetic energy equation

4) kinetic enrgy equation is  0.5 times m times v squared. i re - arranged so speed equals EK DIVIDED by m times 0.5.

5) to work this out i did 0.44 divided by 0.4 times 0.1 which gave me 0.088 m/s. thats the answer i got, is it right??

[studiot]

4 hours ago, sahra said:

idk if my answer is right, could someone please be kind enough to tell me if its right?

the question is A 0.1 KG TOY CONTAINS A COMPRESSED SPRING. WHEN THE SPRING IS RELEASES, THE TOY FLIES 0.45M UPWARDS FROM GROUND LEVEL, BEFORE FALLING BACK DOWN TO THE GROUND. ASSUMING THERE IS NO AIR RESISITANCE, CALCULATE THE SPEED OF THE TOY WHEN IT HITS THE GROUND. GRAVATIONAL FIELD STRENGH IS 9.8 N/KG. here's my working out...

1) i figured out the gravational potential energy = to the kinetic energy in this question

2)gravational potential energy formula is  = mgh so i did o.1 times 0.45 times 9.8 which gave me the answer 0.44j

3) i now figured  i need to work out the speed of the toy meaning i need to re- arrange the kinetic energy equation

4) kinetic enrgy equation is  0.5 times m times v squared. i re - arranged so speed equals EK DIVIDED by m times 0.5.

5) to work this out i did 0.44 divided by 0.4 times 0.1 which gave me 0.088 m/s. thats the answer i got, is it right??

1) Not quite - it is the change in gravitational potential. All all times and positions the toy has some gravitational potential.

2) OK so given the above why does h = 0.45?

3) Yes

4) EK should be conventionally KE, but otherwise yes if you put in the missing square root 

5) 0.5mv2 = mgh  or v = sqrt(2gh).  why did you bother to work out the actual energy and not simple cancel the mass?

6) I think when you recalculate the arithmetic you will find the answer closer to 3m/s

[studiot]

Final thought.

 

I have worked through discussing your method.

 

Why did you choose this method?

It is not the only way to answer this problem, there are others, one famously due to Galileo.

 

It is worth understanding the other, simpler methods first.