Question

Homework for Lab 15: Simple Harmonic Motion Name Date Section 15 10 -5 -15-10-5 0 5 10 15 Displacement (em) Figure 15.6: Force vs displacement graph for a 0.75 kg cart on a horizontal spring. 1. Figure 15.6 shows the force exerted by the spring on a 0.75 kg cart, as a function of its displacement from equilibrium. Positive displacements (in cm) represent stretching of the spring; negative displacements represent compression. Find the spring constant. 2. The cart is moved 2.0 cm beyond the equilibrium point (the spring is stretched). 259

Answer 1

The expression for the restoring force developed in the spring is as follows: F=-kAx From the above equation, the expression for the force constant is as follows: F k=- Ax From the graph, the slop of the curve gives the spring constant. Here, the slope is negative. But the spring That is, the spring constant is as follows: k=Slope ΔΥ -(-10 N-ON) 15x10-2 m-0 m = 66.67 N/m

From the given data, the cart is moved 2.0 cm from its equilibrium point. That is, x =2.0 cm Hence, the elastic potential energy stored in the spring is as follows: = }(66.67 N/m)(2.0x102 m) = 0.0133 J From the expression U = = kx , the energy doesn't depend on the mass of the cart. 2 (b) When the cart is released from rest at 2.0 cm, it passes through equilibrium position. That is, the total elastic potential energy is converted into total kinetic energy of the cart. So that, the kinetic energy of the cart is as follows: K=U = (66.67 N/m)(2.0x10-2 m)? = 0.0133J

The expression for the kinetic energy is as follows: KE = mv Hence, the velocity of the cart is as follows: 2(0.0133J) V 0.75 kg = 0.188 m/s