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**Unformatted text preview: **Using F = ma on both objects, Equation (2) becomes T = M 1 a and Equation (3) becomes m 2 g − T = m 2 a Combining the two equations by substituting T in equation (3) 1 It becomes m 2 g − M 1 a = m 2 a Further; m 2 g = (M 1 + m 2 ) a a = m 2 g / (M 1 + m 2 ) = (9.00 kg · 9.80 m/s 2 )/ (5.00 kg + 9.00 kg) =88.2/14= 6.30 m/s 2 c) Find the tension in the string; From eq. (2) T = M 1 a, but a = m 2 g / (M 1 + m 2 ) Hence T= (M 1 m 2 g)/ (M 1 + m 2 ) = (5.00 kg · 9.00 kg · 9.80 m/s 2 )/ (5.00 kg + 9.00 kg) = (441/14) N =31.5 N 3. Use the (velocity vs time) graph below to answer the following three questions: (2points) a) What is acceleration of an object and its error according to the graph? 4.393+/-0.3126m/s 2 b) What physics quantity does the slope of the velocity vs time graph presents? It presents acceleration c) If the graph was generated using the set-up of Lab “Newton’s Second Law” was the cart moving toward or away from the motion sensor? The cart was moving away from the motion sensor. It is because the graph is on the positive side of x-axis. 2...

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- Spring '15
- Force, net force, Newton’s Second Law