Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: R

There are no items in this cart.
Continue Shopping
Aditya Narayanan C Grade: 12
        Two blocks 2kg & 4kg are connected through a massless string. Coeff. of friction btw 2kg & ground is 0.4 & coeff. of friction btw 4kg & ground is 0.6. F1=10N is applied on 2kg block to the left and F2=20N is applied on 4kg block to the right. Calculate frictional force btw 4kg block and ground. (assume tension is zero before F1 and F2 is applied).Pls solve it.
7 years ago

Answers : (2)

Badiuddin askIITians.ismu Expert
147 Points

Dear Aditya

maximum friction force that can act in between 2 kg block and ground is =0.2*20 = 4

maximum friction force that can act in between 4 kg block and ground is =0.6*40 = 24

when 2kg block just start moving toward left

  equation on 2 kg block  

so 10 -T- 4 =0

      T =6

equation on 4 kg block

  calculate 20-T = 14

which is less than the max friction force for 4 kg block so it will remain in the state of rest

 and frictional force is 14 .

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
Askiitians Experts

7 years ago
Aditya Narayanan C
8 Points

i got the answer as 18 N. Briefly explain it.

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details