The P.E. of a spring, when stretched through 10 c.m. is 5J. The work done to stretch the spring through an additional distance of 10 c.m. is
Answers
Answered by
3
Explanation:
P.E = 5j
x1 = 10cm
X = 10 + 10 = 20cm
x2 = 10 cm
W = ?
P.E = 1/2k(x1)²
= 1/2k(10)²
= 1/2k(100) ---(1)
W = 1/2k(X)²-(x1)²
= 1/2k(20)²-(10)²
= 1/2k(300)j
= 3 × 1/2k(100)
W = 3j ---> 3joule is needed to stretch extra 10cm from existing state ....:);)
plz mark as BRANIEST
Similar questions