Using Center of Mass for Complex Motion
You've plotted and analyzed the Legs measurement for the snowboarder, but that may not be the most representative way to track the snowboarder's position, velocity, and acceleration. He tucks his legs at the peak of the jump and then stretches them out below and in front of his path before he lands. He makes similar wild changes in position with his head, arms, and torso. No matter how his "parts" flail around, though, his center of mass should follow a projectile path. Tracker allows you to mathematically plot a center of mass point from other points on an object, even an object as complex as a human body.
Answers
Answer:
Question :-
✯ If a force of 12 N acting on a body changes its velocity uniformly from 2 m/s to 5 m/s in 20 s then what is the mass of the body?
Given :-
✯ If a force of 12 N acting on a body changes its velocity uniformly from 2 m/s to 5 m/s in 20 s.
Find Out :-
✯ Then what is the mass of the body?
Solution :-
We have :
Force = 12 N
Initial velocity u = 2 m/s
Final velocity v = 5 m/s
Time = 20 s
✭ In case of acceleration :-
As we know that :
☆ \red{ \boxed{\sf{v =\: u + at}}}
v=u+at
☆
So, according to the question or ATQ :-
\leadsto \sf 5 =\: 2 + a(20)⇝5=2+a(20)
\leadsto \sf 5 - 2 =\: 20a⇝5−2=20a
\leadsto \sf 3 =\: 20a⇝3=20a
\leadsto \sf \dfrac{3}{20} =\: a⇝
20
3
=a
\leadsto \sf\boxed{\bold{\red{a =\: \dfrac{3}{20}\: m/s^2}}}⇝
a=
20
3
m/s
2
✭ In case of mass :-
As we know that :
★ \red{ \boxed{\sf{F =\: ma}}}
F=ma
★
So, according to the question or ATQ :-
\leadsto \sf 12 =\: m \times \dfrac{3}{20}⇝12=m×
20
3
\leadsto \sf \dfrac{12}{\dfrac{3}{20}} =\: m⇝
20
3
12
=m
\leadsto \sf \dfrac{12}{1} \times \dfrac{20}{3} =\: m⇝
1
12
×
3
20
=m
\leadsto \sf \dfrac{240}{3} =\: m⇝
3
240
=m
\leadsto⇝ {\small{\bold{\purple{\underline{m =\: 80\: kg}}}}}
m=80kg
Henceforth, the mass of the body is 80 kg.
Correct options is C) 80