A 6 kg block is released from rest at the top of an incline, as shown above, and slides to the bottom. The incline is 1.97 m lon
1 answer:
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Answer:
(a) = 4.19 x
N, and its direction is towards
.
(b) It must be placed inside a hollow shell.
Explanation:
Let, = 165 kg,
= 465 kg,
= 60 kg, and the distance between
and
is 0.340 m.
(a) Since is placed midway between
and
, then its distance to both masses is 0.170 m.
From the Newton's law of universal gravitation,
F =
Where all variables have their usual meaning.
Then,
a. =
-
=
= 2.25 x N
=
= 6.44 x N
∴ = = 6.44 x
- 2.25 x
= 4.19 x N
The net force exerted by the two masses on the 60 kg object is 4.19 x N.
(ii) // = /
/ - /
/
= 6.44 x - 2.25 x
= 4.19 x N
(iii) The direction of the net force is to the right i.e towards .
(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.