A force of 5 N is applied on a 20 Kg mass at rest. The work done in the third second is
Answers
Answered by
47
F=5N
m=20 kg
t=3s
A=f/m
a=5/20=1/4 m/s^2
t=3 sec
u=0 m/s
s=ut+1/2ut^2
1/2*2.25=1.125m
F=5*1.125=5.625
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m=20 kg
t=3s
A=f/m
a=5/20=1/4 m/s^2
t=3 sec
u=0 m/s
s=ut+1/2ut^2
1/2*2.25=1.125m
F=5*1.125=5.625
hope u get your ans plz like and make brainlist
victory1venkatesh:
Sorry bro. Its wrong
why
t = 3 sec = 180 sec. What are you talking
3×60=180
bro
ya vo meri mistake hai but baki dono ka accerilation to same tha
sorry for u do bar lik diya hai vo 1/2at² hai
Answered by
20
Answer:
25/8 joules
Sn = u+a/2(2n - 1)
= 0+1/8(6-1)
= 5/8 m
W = F.S
= 5/8 × 5
25/8 J
hope this helps you
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