The distance(S)-time(t) graph of an object of mass 2 kg moving along a straight line is as shown below. The net work done on the object in the first four seconds is
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Answered by
3
Answer: 8J is the right answer
Explanation:
there are two methods to solve this problem:
1] this is by using simple math:
v=s/t
=8/4
=2m/s
a=v/t
=2/4
=0.5 m/s^2
work=f*s
=m*a*s
=2*0.5*8
=8J
2] By using calculus
ds/dt=v
d[2t]/dt
2=v
dv/dt=a
2/4=a
0.5=a
work=m*a*s
=2*0.5*8
=8J
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Answered by
0
Answer:
v=s/t
=8/4
=2m/s
a=v/t
=2/4
=0.5 m/s^2
work=f*s
=m*a*s
=2*0.5*8
=8J
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