A particle of mass 0.5 kg is displaced from position r1(2 3 1)
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Answered by
0
Answer:
30✓3 J this is an answer of your question
Explanation:
please give this answer brilliantGiven that,
m=0.5kg
r
1
=(2,3,1)
r
2
=(4,3,2)
F=30 N
F
=
i
^
+
j
^
+
k
^
The work done is the dot product of the force and displacement.
The vector
r
2
r
1
=r
2
−r
1
[
4
3
2
]−[
2
3
1
]=[
2
0
1
]
The magnitude of the force
∣
F
∣=30N
The unit vector
f
=
(1)
2
+(1)
2
+(1)
2
1
×
⎝
⎜
⎜
⎛
1
1
1
⎠
⎟
⎟
⎞
=
⎝
⎜
⎜
⎜
⎜
⎜
⎜
⎛
3
1
3
1
3
1
⎠
⎟
⎟
⎟
⎟
⎟
⎟
⎞
Now, the force
F
=F×
f
^
F
=
∣
∣
∣
∣
F
∣
∣
∣
∣
⋅
f
^
=
⎝
⎜
⎜
⎜
⎜
⎜
⎛
3
30
3
30
3
30
⎠
⎟
⎟
⎟
⎟
⎟
⎞
Now, the work done is
W=
F
⋅(r
2
r
1
)
W=
⎝
⎜
⎜
⎜
⎜
⎜
⎜
⎛
3
30
3
30
3
30
⎠
⎟
⎟
⎟
⎟
⎟
⎟
⎞
(
2
0
1
)
W=(
3
60
+0+
3
30
)
W=30
3
J
Hence, the work done is 30 ✓3 j
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