A sphere of mass 1 kg rests at one corner of a box- The box is then moved with a velocity v-8t -2t2- 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-

Given 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_a= 8î 6ĵ ⇒ |F_net=10 N| ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

20 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$F_{n e t} = F_{p} + F_{a} = - 8^i - 6^j$
$\Rightarrow | F_{n e t} = 10 N |$

Suggest Corrections

Similar questions

Q. 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]

Q. A sphere of mass

1 k g

rests at one corner of a cube. The cube is 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 cube at

t = 1

is (

g = 10 m s^{- 2}

) [Figure shows vertical plane of the cube]:

Q. A 1.5 kg box is initially at rest on a horizontal surface when at t =0 a horizontal force F = (1.8t) iN (With t in seconds), is applied to the box. The seconds), is applied to the box. The acceleration of the box as a function of time t is given by.
(g =

10 m / s^{2}

)

¯ a = 0 f o r 0 \leq t \leq 2.85

¯ a = (1.2 t - 2.4) ¯ j m / s^{2} f o r t > 2.85

The coefficient of kinetic friction between the box and the surface is

Q. A box of mass

20 k g

is moving with a velocity of

2 m s^{- 1}

on a rough floor. The box moves

5 m

on the floor before coming to rest. What must be the frictional force acting on the box?

Q. A horizontal force acting on a box of mass 1

k g

has a magnitude of

F = 1 t^{2} N

, where

t

is in second. If the box starts from rest, determine its speed at

t = 3 s

. The coefficient of static friction and kinetic friction between the box and horizontal surface are

μ_{s} = 0.4

and

μ_{k} = 0.3

respectively.

(g = 10 m / s^{2})

Join BYJU'S Learning Program

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]