SOS!! I have this physics problem to solve and im not sure how?
A scale at a grocery store is made of a metal pan (of negligible mass) placed on an ideal spring. The
spring has a force constant of 655 N/m. A customer gently rests 1.00 kgs of bananas and 2.00 kg of
potatoes on the scale. When the 2.00 kg potatoes is suddenly removed, how high above the starting
position do the bananas reach?
- WhomeLv 71 month agoFavourite answer
Where is the "starting position"? Is it the elevation of the unloaded scale? The elevation of the scale with bananas only? The elevation of the scale with both bananas and potatoes?
If I assume an origin at the unloaded scale level, adding the bananas brings the scale down.
x = mg/k = 1.00(9.81)/655 = 0.014977... ≈ 1.5 cm
adding the potatoes brings the scale down below the unloaded origin at
x = mg/k = 3.00(9.81)/655 = 0.0449312... ≈ 4.5 cm
The spring now has potential energy of
PS = ½kx² = ½(655)0.045² = 0.6632 J
When released, this spring energy will convert to gravity potential energy in the bananas.
PS = mgh
0.6632 = 1.00(9.81)h
h = 0.06760... ≈ 6.8 cm
So the banana will reach a point
6.8 cm above the fully loaded scale level
or 6.8 - (4.5 - 1.5) = 3.8 cm above the banana loaded scale
or 6.8 - 4.5 = 2.3 cm above the unloaded scale level.
- oubaasLv 71 month ago
x = (2+1)*g/k = 9.807*3/655 = 0.045 m
k/2*x^2 = m*g*h
h = k/2*x^2 / (m*g) = 655/2*0.045^2 /(1*9.807) = 0.068 m (6.8 cm)
- Andrew SmithLv 71 month ago
When you add both masses the spring is compressed and gains energy. When you remove the potatoes ALL that energy goes into the bananas. The bananas stop when the energy is all in potential. Find the lowest point. ( 0.03m) find the energy in the spring at that.
Now find how high above that point the energy will take the bananas. Finally remember to subtract the 0.03 m to get the height above the original position.