Using the simulation , considerthese four cases. Case 1: Mass = 3 kg; Force 1 =

Question

Using the simulation, considerthese four cases.

Case 1: Mass = 3 kg; Force 1 = 9.9 N; Force 2 = 4 N; Force 3 =-4 N.
Case 2: Mass = 2.5 kg; Force 1 = 9.9 N; Force 2 = 0; Force 3=0.
Case 3: Mass = 1.5 kg; Force 1 = 8.9 N; Force 2 = 5 N; Force 3 = -4N.
Case 4: Mass = 1.5 kg; Force 1 = 4.95 N; Force 2 = 0; Force 3 =0.

(a) Rank these cases by the net force on theblock, from largest to smallest. For all parts of this questions weuse only > and/or = signs, e.g. 3>2=4>1 means that incases 2 and 4 the net forces (acceleration, distance) are equal,they are greater than in case 1 and less than in case 3.
1=2=3>43>2>4>11>2>3=41=2>3=4

(b) Rank these cases by the acceleration of theblock, from largest to smallest.
3>2>1=41>2>3=44>2>3>13=2>1=4

(c) Rank these cases by the distance traveledby the block in 3 seconds, from largest to smallest.
1=2=3>41>2>3>43=4>2=13>2>1=4

Explanation / Answer

(a)

Fnet2 = 9.9N + 0N + 0N = 9.9N
Fnet3 = 8.9N + 5N - 4N = 9.9N
Fnet4 = 4.95N + 0N + 0N = 4.95N
Therefore for the forces 1 = 2 = 3 >4
(b) Now a1 = Fnet1/m1 = 9.9N/3kg = 3.3m/s^2
a2 = Fnet2/m2 =9.9N/2.5kg = 3.96m/s^2
a3 = Fnet3/m3 =9.9N/1.5kg = 6.6m/s^2
a4 = Fnet4/m4 =4.95N/1.5kg = 3.3m/s^2
Therefore for the acceleration 3 > 2 > 1 = 4
(c)
We have the distance traveled is x = (1/2)at2

Then x1 = (0.5)(3.3m/s^2)(3s)2 = 14.85m
x2 = (0.5)(3.96m/s^2)(3s)2 =17.82m
x3 = (0.5)(6.6m/s^2)(3s)2 =29.7m
x4 = (0.5)(3.3m/s^2)(3s)2 =14.85m
Therefore for the distance traveled 3 > 2 > 1 = 4

Fnet2 = 9.9N + 0N + 0N = 9.9N
Fnet3 = 8.9N + 5N - 4N = 9.9N
Fnet4 = 4.95N + 0N + 0N = 4.95N
Therefore for the forces 1 = 2 = 3 >4
(b) Now a1 = Fnet1/m1 = 9.9N/3kg = 3.3m/s^2
a2 = Fnet2/m2 =9.9N/2.5kg = 3.96m/s^2
a3 = Fnet3/m3 =9.9N/1.5kg = 6.6m/s^2
a4 = Fnet4/m4 =4.95N/1.5kg = 3.3m/s^2
Therefore for the acceleration 3 > 2 > 1 = 4
(c)
We have the distance traveled is x = (1/2)at2

Then x1 = (0.5)(3.3m/s^2)(3s)2 = 14.85m
x2 = (0.5)(3.96m/s^2)(3s)2 =17.82m
x3 = (0.5)(6.6m/s^2)(3s)2 =29.7m
x4 = (0.5)(3.3m/s^2)(3s)2 =14.85m
Therefore for the distance traveled 3 > 2 > 1 = 4

Fnet3 = 8.9N + 5N - 4N = 9.9N
Fnet4 = 4.95N + 0N + 0N = 4.95N
Therefore for the forces 1 = 2 = 3 >4
(b) Now a1 = Fnet1/m1 = 9.9N/3kg = 3.3m/s^2
a2 = Fnet2/m2 =9.9N/2.5kg = 3.96m/s^2
a3 = Fnet3/m3 =9.9N/1.5kg = 6.6m/s^2
a4 = Fnet4/m4 =4.95N/1.5kg = 3.3m/s^2
Therefore for the acceleration 3 > 2 > 1 = 4
(c)
We have the distance traveled is x = (1/2)at2

Then x1 = (0.5)(3.3m/s^2)(3s)2 = 14.85m
x2 = (0.5)(3.96m/s^2)(3s)2 =17.82m
x3 = (0.5)(6.6m/s^2)(3s)2 =29.7m
x4 = (0.5)(3.3m/s^2)(3s)2 =14.85m
Therefore for the distance traveled 3 > 2 > 1 = 4

Fnet4 = 4.95N + 0N + 0N = 4.95N
Therefore for the forces 1 = 2 = 3 >4
(b) Now a1 = Fnet1/m1 = 9.9N/3kg = 3.3m/s^2
a2 = Fnet2/m2 =9.9N/2.5kg = 3.96m/s^2
a3 = Fnet3/m3 =9.9N/1.5kg = 6.6m/s^2
a4 = Fnet4/m4 =4.95N/1.5kg = 3.3m/s^2
Therefore for the acceleration 3 > 2 > 1 = 4
(c)
We have the distance traveled is x = (1/2)at2

Then x1 = (0.5)(3.3m/s^2)(3s)2 = 14.85m
x2 = (0.5)(3.96m/s^2)(3s)2 =17.82m
x3 = (0.5)(6.6m/s^2)(3s)2 =29.7m
x4 = (0.5)(3.3m/s^2)(3s)2 =14.85m
Therefore for the distance traveled 3 > 2 > 1 = 4
Therefore for the distance traveled 3 > 2 > 1 = 4

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