A sphere of mass 1 kg rests at one corner of a box- The box is then moved with a velocity v -8 t -2 t 2- Where t is time in second- The force by sphere on the box at t -1 s is g -10 ms -2 -Figure shows vertical plane of the box-y-A- 10 NB- 20 NC- 6 ND- 8 N

The correct option is A 10 NGiven v=8tî 2t^2ĵ a=dvdt=8î 4tĵ a|_t=1=8î 4ĵ Pseudo force F_p= ma= 8î+4ĵ Force due to gravity F_g= mgĵ= 10ĵ Now nbsp;F_net=F_p+F ...

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Byju's Answer

Standard XII

Physics

Pseudo Force

A sphere of m...

Question

A sphere of mass $1 kg$ rests at one corner of a box. The box is then moved with a velocity $\to v = 8 t^i - 2 t^{2}^j$ . Where $t$ is time in second. The force by sphere on the box at $t = 1 s$ is ( $g = 10 {ms}^{- 2}$ ) [Figure shows vertical plane of the box]

8 N

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10 N

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20 N

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6 N

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Solution

The correct option is B $10 N$
Given $v = 8 t^i - 2 t^{2}^j$
$\to a = \frac{d v}{d t} = 8^i - 4 t^j$
$\to a |_{t = 1} = 8^i - 4^j$
Pseudo force $F_{p} = - m \to a = - 8^i + 4^j$
Force due to gravity $F_{g} = - m g^j = - 10^j$
Now
$F_{n e t} = F_{p} + F_{a} = - 8^i - 6^j$
$\Rightarrow | F_{n e t} = 10 N |$

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A sphere of mass 1 kg rests at one corner of a box. The box is then moved with a velocity →v=8t ^i−2t2^j. Where t is time in second. The force by sphere on the box at t=1 s is (g=10 ms−2) [Figure shows vertical plane of the box]

The correct option is B 10 NGiven v=8t^i−2t2^j →a=dvdt=8^i−4t^j →a|t=1=8^i−4^j Pseudo force Fp=−m→a=−8^i+4^j Force due to gravity Fg=−mg^j=−10^j Now Fnet=Fp+Fa=−8^i−6^j ⇒|Fnet=10 N|

A sphere of mass $1 kg$ rests at one corner of a box. The box is then moved with a velocity $\to v = 8 t^i - 2 t^{2}^j$ . Where $t$ is time in second. The force by sphere on the box at $t = 1 s$ is ( $g = 10 {ms}^{- 2}$ ) [Figure shows vertical plane of the box]