2 किग्रा द्रव्यमान का एक पिण्ड चिकने क्षैतिज तल पर 10 मीटर सैकण्ड के प्रारम्भिक वेग से पूर्व दिशा में गतिमान है पिण्ड पर 10 न्यूटन का एक समान बल उत्तर दिशा में कार्य करता है, 2 सेकण्ड बाद पिण्ड का अन्तिम वेग होगा:
Answers
Answer:
The mass of body, m=0.40kg
Constant initial speed of the body is, u=10m/s
Force that acts on the body is, F=8.0N
(i)
At t= –5 s
The force starts acting on the body from t=0 s.
So, the acceleration of the body during this time was zero and it moves at a constant speed.
Position of the body is given by:
x=v×t
x=10×(−5)
=−50 m
(ii)
At t=25 s
Since the force acts in the opposite direction of motion of particle. So, the acceleration of the body due to the force acting on it is:
a=
M
F
=
0.40
−8.0
=−20ms
−2
The position is given by the second equation of motion:
s
′
=ut
′
+
2
1
a"t
2
=10×25+
2
1
×(−20)×25
2
=250−6250=−6000m
(iii)
At t=100 s
For first 30 sec, it moves under retardation of force and after that the body moves at constant speed.
a=−20ms
−2
u=10m/s
The distance covered in 30 s:
s
1
=ut+
2
1
a"t
2
=10×30+
2
1
×(−20)×30
2
=300−9000=−8700m
Now, the speed after 30 sec is:
v=u+at
v=10−(20×30)
=−590 m/s
Distance covered in remaining 70 sec:
x
′
=−590×70
=−41300 m
So, total distance covered in 100 s will be:
X=−8700−41300
X=−50000 m=−50 km
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