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Question: A smooth table is placed horizontally and an ideal spring of spring constant \(k = 1000\mspace{6mu} ...

A smooth table is placed horizontally and an ideal spring of spring constant k=10006muN/mk = 1000\mspace{6mu} N/m and unextended length of 0.5m0.5m has one end fixed to its centre. The other end is attached to a mass of 5kg5kg which is moving in a circle with constant speed 20m/s20m/s. Then the tension in the spring and the extension of this spring beyond its normal length are

A

5006muN,6mu0.56mum500\mspace{6mu} N,\mspace{6mu} 0.5\mspace{6mu} m

B

6006muN,6mu0.66mum600\mspace{6mu} N,\mspace{6mu} 0.6\mspace{6mu} m

C

7006muN,6mu0.76mum700\mspace{6mu} N,\mspace{6mu} 0.7\mspace{6mu} m

D

8006muN,6mu0.86mum800\mspace{6mu} N,\mspace{6mu} 0.8\mspace{6mu} m

Answer

5006muN,6mu0.56mum500\mspace{6mu} N,\mspace{6mu} 0.5\mspace{6mu} m

Explanation

Solution

k=1000,m=5kg,l=0.5m,v=20m/sk = 1000,m = 5kg,l = 0.5m,v = 20m/s (given)

Restoring force = kx =mv2r=mv2l+x= \frac{mv^{2}}{r} = \frac{mv^{2}}{l + x}1000x=5(20)20.5+x1000x = \frac{5(20)^{2}}{0.5 + x}x=0.5mx = 0.5m

and Tension in the spring = kx =1000×12=500N= 1000 \times \frac{1}{2} = 500N