Question

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 $g=10m{s}^{-2})$.

• A. 20 N
• B. 10 N
• C. 60 N
• D. 40 N

Answer 1

The correct option is C 60 N

Take the two masses + spring as a system. The key is to guess the position where the Normal force between the ground and the spring mass system is maximum and calculate that value.
Notice that since 4kg block is stationary and 1kg mass is moving, the centre of mass also moves accordingly. (In fact, centre of mass executes a simple harmonic motion). Now, there are two forces on the spring mass system:

1. Weight or the gravitational force $W=(4+1)g=50N$
2. Normal force $N$

Hence acceleration $a$ of the centre of mass $=\frac{N-W}{4+1kg}$ upwards. In other words, a is maximum when N is maximum and vice versa. Hence the maximum Normal force point is the one where upward acceleration is maximum i.e. the lowest point in the trajectory.
At the lowest point, acceleration of the 1kg block $a=-{\omega }^{2}A=10m{s}^{-2}$
Hence acceleration of the COM $=10\times 1/(4+1)=2m{s}^{-2}$
which should be equal to $\frac{N-50}{5}$ which gives $N=60N$
Note that the Normal force exerted by the spring-mass system on the ground and the Normal force exerted by the spring mass system on the ground form a newton action-reaction pair and hence are equal.