Given that:
k = 500 n/m,
work (W) = 704 J
spring extension (x) = ?
we know that,
Work = (1/2) k x²
704 = (1/2) × 500 × x²
x = 1.67 m
A spring stretched for 1.67 m distance.
<span>B.Extrinsic motivation </span>
Using Newton's second law of motion:
F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)
Given: Find: Formula: Solve for m:
F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2
a= 200m/s^2
Solution:
Using Eq.2
m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg
Complete question:
Two 10-cm-diameter charged rings face each other, 21.0 cm apart. Both rings are charged to +40.0 nC. What is the electric field strength at the midpoint between the two rings ?
Answer:
The electric field strength at the mid-point between the two rings is zero.
Explanation:
Given;
diameter of each ring, d = 10 cm = 0.1 m
distance between the rings, r = 21.0 cm = 0.21 m
charge of each ring, q = 40 nC = 40 x 10⁻⁹ C
let the midpoint between the two rings = x
The electric field strength at the midpoint between the two rings is given as;

Therefore, the electric field strength at the mid-point between the two rings is zero.