Question

looking for the rest of the answers to pages 5-2, 5-6. (graphs are included at the end)

DATA 722.1 Mass of cart with plastic box mounted on it- Mass constant, resultant force varied 0:78 Total moving mass =_7799-m: Acceleration (in m/s Resultant Force 0.1728 0.196 20 gm wt = nt G-311 gm wt = 0.39Z 40gm wt = 0-$584 Resultant force constant, mass varied Resultant forceo gm wt-9.49 nt Reciprocal of total moving mass (Can be obtained when use Excel for Amt of add'l mass in mass holder in cart Total moving mass (kg) (n'sう (gm) o-SC92 0 None 100 G. 9774 l. 0zch 200 0.4245 300 G. 84 75 400 G. 3357 1.2x0 500 O. 3273 Lab Report 5-1

Answer 1

Slope of graph a v > F_R� From the graph of a v > F_R, we have� y = 0.7675 x + 0.0587� Equ of line i.e.� y = mx + c m rightarrow slope� c rightarrow y-intercept� From the above� m = 0.7675 nt/m/s^2� nt/m/s^2 is the unit of the slope� The reciprocal is m/s^2/nt� slope = nt/m/s^2� 1/slope = 1/nt/m/s^2 �= m/s^2/nt� w.r.t� 1 Newton is unit of Force Force = mass times acceleration� = kg times m/s^2� therefore 1 Nt = kg ms^-2 \r\n From graph, unit of slope is� nt/m/s^2 = kg m/s^2 times s^2/m� = 1 kg unit or kg \r\n From Newton's 2nd�law, we have� Force = rate of change of momentum w.r.t time� momentum = mass times velocity� a change in velocity gives raise to momentum & change in velocity w.r.t time given rise to acceleration� If velocity was a constant� i.e. v = 0 �then Force = 0 as there is