F = kx, so
70/65 the stretch means 70/65 the force
1/13 * 5 = 5/13 or 0.3846N
A force of 5n stretches an elastic material by 65mm, what additional force on 1 stretch the material 70mm assumed the elastic limit is not exceeded
2 answers
Using Hooke's Law: F = kx
65 mm = 0.065m, 70 mm = 0.070 m
k = 5 N/0.065 m = 76.9 N/m
F = 76.9 N/m * 0.070 m - 5 N = 0.383 N
Therefore, the additional force should be about 0.383 N.
65 mm = 0.065m, 70 mm = 0.070 m
k = 5 N/0.065 m = 76.9 N/m
F = 76.9 N/m * 0.070 m - 5 N = 0.383 N
Therefore, the additional force should be about 0.383 N.