Work Energy And Power Question 280

Question: A block lying on a smooth surface with spring connected to it is pulled by an external force as shown. Initially the velocity of ends A and B of the spring are 4 m/s and 2 m/s respectively. If the energy of the spring is increasing at the rate of 20 J/sec, then the stretch in the spring is Critical Thinking

Options:

A) 1.0 cm

B) 2.0 cm

C) 10 cm

D) 2.0 cm

Show Answer

Answer:

Correct Answer: C

Solution:

[c] Let xAandxB be the position of ends A and B at time (from the block, then stretched length of the spring will be l2=xAxB and so the stretch Δ=21=(xAxB)1(1 natural length of the spring) So, U=12kΔ2=12k[(xAxB)1]2

P=dUdt=12k.2(xAxB1)(dxAdtdxBdt)

P=F(vAvB)F=dxAvAvB

Δ=Fk=P(vAvB)=2042×100

Δ=0.1m=10cm



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