If the length of a #44 cm# spring increases to #58 cm# when a #5 kg# weight is hanging from it, what is the spring's constant?

Answer 1

#K=(49,05)/(0,14)=350,357142857 N/m#

#58-44=14 cm# #Delta x=14* 10^-2=0,14 meters# #F=K*Delta x# #5*9,81=K*0,14# #49,05=K*0,14# #K=(49,05)/(0,14)=350,357142857 N/m#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Axel Acevedo

To find the spring constant, you can use Hooke's Law formula: ( F = -kx ), where ( F ) is the force applied (weight), ( k ) is the spring constant, and ( x ) is the displacement from the equilibrium position.

First, calculate the force applied: ( F = mg ), where ( m ) is the mass (5 kg) and ( g ) is the acceleration due to gravity (approximately 9.8 m/s²).

Next, find the displacement (( x )): ( x = \text{final length} - \text{initial length} ).

Finally, use Hooke's Law to calculate the spring constant (( k )): ( k = -\frac{F}{x} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.