Bathroom scale compression problem

  • Thread starter Thread starter shrtweez13
  • Start date Start date
  • Tags Tags
    Compression Scale
AI Thread Summary
When jumping onto a bathroom scale from a height of 1.4 m, the gravitational potential energy (mgh) converts into spring potential energy (1/2 kx^2). The spring constant (k) can be calculated using Hooke's law, given the scale compresses 0.50 mm under a weight of 700 N. The total distance fallen includes both the height of the jump and the compression of the spring. While this additional compression may seem significant, it has a minimal effect on the scale reading at the peak. Understanding these energy transformations is crucial for solving the problem accurately.
shrtweez13
Messages
5
Reaction score
0
if you stand on a bathroom scale, the spring inside the scale compresses .50 mm, and it tells you your weight is 700 N. Now if you jump on the scale from a height of 1.4 m, what does the scale read at this peak.


i found an equation to use, i just can't figure out what I'm solving for. I'm using
mgh = 1/2kx^2

i have m (71.4) g (9.8) and h (1.4). what i don't know is k and x. can anyone help please?
 
Physics news on Phys.org
shrtweez13 said:
i found an equation to use, i just can't figure out what I'm solving for. I'm using
mgh = 1/2kx^2
This is a statement of conservation of mechanical energy: the initial gravitational PE is transformed into spring PE.

k is the spring constant, x is the displacement of the spring from its uncompressed position. You can figure out the spring constant from Hooke's law (using the data supplied in the problem set up): F = -kx.
 
Yeah k can be determined from the given data.

One more thing!

When you jump from a ht. of 1.4 m, the gravitational PE lost by you is definitely stored as PE in spring. But You are not actually falling 1.4 m but you are falling 1.4 m PLUS the distance contracted bythe spring.

In this case, it wouldn't make much of a difference(why?)
Anyway, you should know the concept! :smile:
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top