Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

A small block of mass 1kg is placed over a plank of mass 5kg. The length of the plank is 8m. Coefficient of friction between the block and the plank is 0.5 and the ground over which plank is placed is smooth. A constants force F = 50N is applied on the plank in horizontal direction. The time after which the block will separate from the plank is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

a

1s

b

4s

c

8s

d

2s

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} f_{k} = μ_{k} N \\ f_{k} = 0 .5 (10) = 5 N \end{array}$

Question Image

$For 1 kg, \sum F = ma$
$\Rightarrow 5 = a_{1} \Rightarrow a_{1} = 5 m / s^{2}$
$\begin{array}{l} For 5 kg, \sum F = ma \\ \Rightarrow 50 - 5 = 5 a \\ \Rightarrow a_{2} = 9 m / s^{2} \\ a_{rel} = a_{2} - a_{1} \\ = 9 - 5 = 4 m / s^{2} \\ S_{rel} = u_{rel} t + \frac{1}{2} a_{rel} t^{2} \end{array}$
$\begin{array}{l} \Rightarrow + 8 = 0 + \frac{1}{2} (4) t^{2} \\ \Rightarrow t = 2 s \end{array}$

Watch 3-min video & get full concept clarity

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!

A small block of mass 1kg is placed over a plank of mass 5kg. The length of the plank is 8m. Coefficient of friction between the block and the plank is 0.5 and the ground over which plank is placed is smooth. A constants force F = 50N is applied on the plank in horizontal direction. The time after which the block will separate from the plank is