Lab 112: Newton’s Second Law
1. Objective:
2. Theoretical background
3. Procedures
4. Results
a. Experimental Data
Mg=178.83g=0.17883kg, Mg is the mass of glider
Mh=40g=0.04kg, Mh is the mass of a hanging weight, M1=50.28g, or 0.05028kg
X0=90m, L=54.5m, X0 is the initial position of glider, and L is distance between 2 photogates.
ϴ=5o, angle of inclined air track.
Total Glider mass (kg) |
Hanging mass Mh(kg) |
Acceleration (m/s2) |
Time to travel distance L (s) |
Velocity at gate 1 V1 (m/s) |
Velocity at gate 2 V2 (m/s) |
Mg |
0.04 |
1.79 |
0.457 |
0.915 |
1.753 |
Mg+2M1 |
0.04 |
1.279 |
0.549 |
0.768 |
1.486 |
Mg+4M1 |
0.04 |
0.953 |
0.633 |
0.667 |
1.284 |
Mg ϴ=5o |
0.04 |
1.141 |
0.569 |
0.732 |
1.396 |
Mg+2M1 ϴ=5o |
0.04 |
0.522 |
0.845 |
0.486 |
0.938 |
Mg+4M1 ϴ=5o |
0.04 |
0.149 |
1.608 |
0.242 |
0.486 |
b. Calculation
(1)Finding acceleration
a=F/M= Mh(g)/(Mg+Mh). a=(40*9.8)/(178.83+40)=1.79m/s2, we can the proceed to find values of acceleration by adding more mass to the glider.
(2)Finding acceleration at inclined angle
a=F/M=[Mgsin(5o)-Mh]*g / (Mg+Mh), a=[178.83*0.087-40]*(-9.8) / (178.83+40)=1.093m/s2.
(3)Finding velocity
V2=2ax, the v0 is zero and x is displacement from glider to gate 1, x is 0.196m for V1, and 0.741m for V2
V=sqrt[2*1.79*0.196]=0.838m/s
(4)Finding time
T=(V2-V1)/a, T=(1.629-0.838)/1.79=0.442s
Total Glider mass (kg) |
Hanging mass M(kg) |
Acceleration (m/s2) |
Time to travel distance L (s) |
Velocity at gate 1 V1 (m/s) |
Velocity at gate 2 V2 (m/s) |
Mg |
0.04 |
1.79 |
0.442 |
0.838 |
1.629 |
Mg+2M1 |
0.04 |
1.227 |
0.534 |
0.694 |
1.347 |
Mg+4M1 |
0.04 |
0.933 |
0.612 |
0.605 |
1.176 |
Mg ϴ=5o |
0.04 |
1.093 |
0.565 |
0.656 |
1.272 |
Mg+2M1 ϴ=5o |
0.04 |
0.48 |
0.853 |
0.434 |
0.844 |
Mg+4M1 ϴ=5o |
0.04 |
0.161 |
1.475 |
0.251 |
0.488 |
c. Error analysis
Mg (g) |
Pulse(s) |
V1(m/s) |
V2(m/s) |
a (m/s2) |
Pulse(s) |
V1(m/s) |
V2 (m/s) |
a(m/s2) |
178.83 |
.0150 |
.0770 |
.1237 |
.0020 |
.0043 |
.0773 |
.1240 |
.0477 |
279.39 |
.0153 |
.0744 |
.1373 |
.0517 |
.0083 |
.0521 |
.0944 |
.0418 |
379.95 |
.0210 |
.0620 |
.1078 |
.0196 |
.1329 |
.0090 |
.0020 |
.0117 |
178.83 |
3.39% |
9.19% |
7.59% |
.11% |
.76% |
11.81% |
9.75% |
4.36% |
279.39 |
2.87% |
10.73% |
10.18% |
4.21% |
.97% |
12.02% |
11.19% |
8.70% |
379.95 |
3.43% |
10.25% |
9.17% |
2.10% |
9.01% |
3.57% |
.41% |
7.26% |
5. . Discussion:
The experiment consists of 2 parts, with second part of experiment an inclined angle. The measured data allows us to find the acceleration of the glider with various masses using newton’s second law, F=ma. We can then proceed to calculate other missing values, velocity and time, from the experiment using equations of motion. There are errors produced as referred to table I and table II. If calculation were done correctly, then % difference could very well tell us something had gone wrong during the experiment. Factors such negligence of flag and cord’s mass, inaccurate measurements could result in error.
6. Conclusion
The newton’s second law of motion establishes the relationship between mass and acceleration, as their product gives the net force acting on an object. Acceleration is therefore inversely proportion to the mass of the object. In this experiment, newton’s law is well demonstrated as increase in mass leads to decrease in acceleration, or vise versa.
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