A constant force acts on an object of mass 2kg for 10s and increases its velocity from 5m/s
to 10m/s. Find the magnitude of force applied.
Answers
Answered by
2
Answer:
Step-by-step explanation:
M=2kg=
t=10sec
u=5m/sec
v=10m/sec
v=u+at
a=(v-u)/t=(10-5)/10=1/2m/sec²
F=Ma=2×1/2=1kgm/sec²=1N
if this force is applied for 15sec then, velocity is
F=mv/t
v=(1×15)/2=7.5m/sec.
Answered by
2
Answer:
1N
Step-by-step explanation:
M=2kg=
t=10sec
u=5m/sec
v=10m/sec
v=u+at
a=(v-u)/t
=(10-5)/10=
1/2m/sec²
F=Ma
=2×1/2
=1kgm/sec²
=1N
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