an objectof mass 2 kg falls from height of 10m into the sand.the object moves 2cm before coming to rest... what is the average resistive force
And also explain what is resistive force... Not from google..
Answers
Answered by
9
When the body drops from height 2m & will reach the ground, its u = 0
v = ?
v2 - u2 = 2gh
v2 = 2 * 9.8 * 2 = 4 * 9.8
Now, this v is the initial velocity for calculation retardation
And final velocity is zero
And distnace s = 10 cm
from the formula v2 - u2 = 2as
a = v2-u2 /2s
a = -196 m/s2
v = ?
v2 - u2 = 2gh
v2 = 2 * 9.8 * 2 = 4 * 9.8
Now, this v is the initial velocity for calculation retardation
And final velocity is zero
And distnace s = 10 cm
from the formula v2 - u2 = 2as
a = v2-u2 /2s
a = -196 m/s2
Answered by
2
Answer:
4900N
Explanation:
Potential energy= kinetic energy = 1/2mv2
or, mgh=1/2mv2
V= √2gh =√2×9.8×10 = 14 m/s
object will pass through sand and loos Ek due to friction
hence unknown mass of sand m=0
Fs = 1/2 mv2
= 1/2 ×14×14
=98
s = 2cm= 0.02m
F= 98/0.02
= 4900N
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