JEE Main 2014 · 9 Apr · Shift 1 · Physics Q5

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PhysicsSingle correct+4 / -1Question 5 of 90

Two bodies of masses 1 kg and 4 kg are connected to a vertical spring, as shown in the figure. The smaller mass executes simple harmonic motion of angular frequency 25 rad/s, and amplitude 1.6 cm while the bigger mass remains stationary on the ground. The maximum force exerted by the system on the floor is (take $g=10$ ms $^{-2}$ ).

A 4 kg block on the floor with a vertical spring on top of it carrying a 1 kg block.

20 N

10 N

60 N

40 N

Answer and worked solution

Official answer: C

The floor force is largest when the spring is most compressed, at the lowest point of the 1 kg mass:
$N_{\max}=(m_1+m_2)g+m_1\omega^2A=50+1\times625\times0.016=50+10=60\text{ N}.$

Topics:Spring-mass systems Simple harmonic motion

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 5