I dont know where to go after a how do a calculate Wm Mt and (Mt)a Question Deta

Question

I dont know where to go after a how do a calculate Wm Mt and (Mt)a
Question Details
For this SLP you will experiment with a simulated wagon, or sled (Background: Fendt, 2003). The experiment is set up so that you can change the mass of the sled, the value of the hanging mass, and the coefficient of friction between the sled and the surface supporting it.

The Experiment





The mass of the wagon (M) and the hanging mass (m) are in kilograms. The total mass of the system is Mt, the sum of M and m. The acceleration of gravity (g) is 9.8 m/s2 . The acceleration of the system (a) varies, depending upon the experimental values of M, m, and µ. In the situation shown above (µ = 0), the only force acting on the system is the weight of the hanging mass, Wm.

Fill in the following table, using the value of a obtained from the simulation. Record and calculate data to the nearest three decimal places. The masses in the table are in kg, but the input data are in grams, so make the necessary conversions. The first row has been completed for you.

M m a Wm Mt (Mt)a

.100 .01 .892 .098 .11 .098

.100 .03 2.264

.100 .05 3.270

.100 .07 4.039

.100 .10 4.905


I just need to find out how to complete the second row and I can go from there?Just provide me with a step by step for the second row and I can finish it up?

Explanation / Answer

mass of wagon = M, hanging mass = m, acc. of system = a, weight oh hanging mass =Wm Mt = total mass M, m, a are given in all rows onwards second row we have to find out Wm = m*g Mt= M+m (Mt)a = (M+m)*a for second row M=.100 m= .03 a = 2.264 Wm=0.100 * 9.8 = .98 Mt = .100+ .03 = .13 (Mt)a = (.100 + .03) *2.264 = .29432 so second row i s .100 .03 2.264 .98 .13 .29432

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