Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
(a) 0.71 mm
(b) 0.158 cubic cm
Explanation:
The width of one wire is the diameter of the wire.
(a) Let the diameter of each wire is d.
So, 10 d = 14.2 mm
d = 1.42 mm
radius of each wire, r = d/2 = 1.42/2 = 0.71 mm
(b) Length, L = 10 cm
The volume of the single wire is given by
Data Analysis and Conclusion