Hooks Law Lab

Ans: Increasing the weight would increase the stretch of the spring. Since a larger weight will pull a spring with a greater force, the stretch of the spring will increase. Ans: The period of the spring does not change even if the weight is increased. Increasing the weight increases the displacement of the spring from its mean position, but it also increases its acceleration. Hence the spring takes equal time to cover this increased distance and the period remains unchanged even if the weight is increased. Ans: Mass Gram Kg 0 0 50 0.050 100 0.100 150 0.150 200 0.200 250 0.250 300 0.300 The graph is produced below: Extension Weight (N) 0 0.5 1 1.5 2.0 2.5 3.0 cm 0.0 1.0 2.1 3.0 3.9 5.0 6.1 m 0 0.010 0.021 0.030 0.039 0.050 0.061 Ans: Mass Extension Gram Kg Weight (N) cm m 0 50 100 150 200 250 300 0 0.05 0.1 0.15 0.2 0.25 0.3 0 0.5 1 1.5 2 2.5 3 0 1 2.1 3 3.9 5 6.1 0 0.01 0.021 0.03 0.039 0.05 0.061 The required graph is shown below: Work Done by Gravity 0 0.005 0.021 0.045 0.078 0.125 0.183 Table 3 Experimental Time (seconds) Spring Constant, K (N/m) Theoretical Time-period (T) Mass (kg) 0.30 50 0.28 0.100 -7% 0.35 50 0.34 0.150 -2% 0.41 50 0.40 0.200 -3% 0.47 50 0.44 0.250 -6% 0.50 50 0.49 0.300 -3% % Error Observations: The experimental and the theoretical results differ by -7% to -2%. Conclusion: The difference can be attributed to human error is proper data recording. Ans: 1. Restoring Force: This is any one of the forces or torques that tend to restore a system or parts thereof to equilibrium after displacement. 2. Spring Constant: The restoring force in a spring is proportional to the displacement of the spring from its mean position. The proportionality constant of this proportion is known as spring constant. 3. Hooke’s Law: Hooke’s law states that the strain of the material is proportional to the applied stress within the elastic limit of that material. 4. Elongation: Elongation is a measure of deformation that occurs before a material eventually breaks when subjected to a tensile load. As the latter is applied, an increase in length and a uniform reduction in cross-sectional area take place, while the material maintains a constant volume. Elongation due to expansion can also occur when a material undergoes an increase in temperature, or if both an axial force and a high temperature act simultaneously upon a material. Ans: This formula states that the restoring force generated within a spring is directly proportional to its elongation form the equilibrium position. Ans: The restoring force generated within a spring is equal to the product of a constant, called spring constant, and the elongation of the spring. Ans: The elastic potential energy stored within a spring is equal to the half of the product of the spring constant and square of the elongation of the spring. Ans: Force, F , on the spring is: F=mg ,F=20× 10 F=200N The spring’s stretch is 5.0 cm. Hence, from Hooke’s Law, F=−kx 200=−k×0.05 −200 ,k= 0.05 k=−4000N / m Since spring constant is a scalar quantity, ignore the negative sign. The value of the spring constant is 4000 N/m. Ans: 1lb=453.6gm 40lb=40× 453.6gm 40lb=18.144kg Here, F=mg F=18.144×10 F=181.44N F=−kx F=−k×0.3045 −181.44 ,k= 0.3045 (ignoring the sign) k=595.86N / m Ans: Assume the stretch is x . So, F=−kx 28× 10=−60x x= ,x=4.67m −280 60 (ignoring the sign) Ans: F=−kx 60× 10=−k×0.86 600 0.86 k= 697.67N / m k=