A physics student designed an apparatus for her experimental physics class. The

Question

A physics student designed an apparatus for her experimental physics class. The apparatus consisted of a block m = 0.25 kg mass attached to a compressed spring of constant k mounted on a cart of mass M = 2 kg, as shown in Figure 3.6. The cart is pulled by a string that passes over a pulley to a 10 kg hanging weight. The student used photogate timers to determine the acceleration of the mass on the spring when it was 15 cm from its equilibrium position. She found this acceleration to be -2 m/s^2 relative to the lab bench. What is the spring constant? (You may assume there is no friction in the system and the spring is massless.)

Explanation / Answer

Given that when the mass is at 15 cm from equilibrium position it accelerates to -2 m/s^2

Then,F=ma= -kx2

k=- ma/(x2)= -(0.25Kg×-2m/s2)/(15×10-2m)^2

=22.22N/m(spring constant is constant for a given spring)

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