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Unformatted text preview: 1 1 applied T f F M a =-The force applied is equal to the weight of the hanger and is 6.67*9.8=65.37N The total mass is equal to 109 as stated above and the acceleration was found to be .308. 65.37-(109*.308)= 31.79 f = _31.79_N Phys310 Lab Page 2 Experiment 6 Physics 310 Experiment 6 continued 7. Data Table 3 Run Hanging mass (kg) F applied (N)= m H g F net (N)=F applied-f #2 0.0118 #3 0.0175 #4 0.0231 8. Data Table 4 Run Acc., theory (m/s 2 )= F net / M T Acc., exp. (m/s 2 ) % difference #2 #3 #4 Questions 9. Why did the slope change for each run? 10. For runs #2, #3, #4, what did you observe about the slope of the Linear Fit as the net force increased? 11. What are the units for the slope for each graph? Explain. 12. What happens to an objects acceleration if the net force applied to the object is increased but the objects mass remains constant? Phys310 Lab Page 3 Experiment 6...
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