A truck of mass 5000 kg rolls down a hill starting from rest. If it covers 200 m in 10 s then force
acting on it will be
Answers
Answered by
33
m= 5000kg
v=200/10=20m/s
u=0
t=10
f = m×a
=m×((v-u)/t)
=5000×((20-0)/10)
=5000×(2)
=10000N
if this answer is wrong then please inform me!!in the comment box
v=200/10=20m/s
u=0
t=10
f = m×a
=m×((v-u)/t)
=5000×((20-0)/10)
=5000×(2)
=10000N
if this answer is wrong then please inform me!!in the comment box
kundan1234:
Ans is 3 bullets
bullets???
Soory nothing.
it's wrong
Answered by
17
Given :
Initial speed=u= 0 m/s
mass=5000kg
Distance=s=200m
time=10s
Force=?
Formula to be used =F=ma
where a=v-u/t
let us find out final speed
Speed =V=Distance/time
=200/10=20m/s
Now, F=m[v-u]/t
=5000[20-0]/10
=5000x20/10
=10,000N
∴Force acting on truck is 10,000N
Initial speed=u= 0 m/s
mass=5000kg
Distance=s=200m
time=10s
Force=?
Formula to be used =F=ma
where a=v-u/t
let us find out final speed
Speed =V=Distance/time
=200/10=20m/s
Now, F=m[v-u]/t
=5000[20-0]/10
=5000x20/10
=10,000N
∴Force acting on truck is 10,000N
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