An iron block of mass 5 kg is kept on a trolley. If the trolley is being pushed with an acceleration of 5 m / s 2, what will be the force of friction between the block and the trolley surface? (Take the coefficient of static friction between the block and the trolley surface to be 0,8).

Question

An iron block of mass 5 kg is kept on a trolley. If the trolley is being pushed with an acceleration of 5 m / s 2, what will be the force of friction between the block and the trolley surface? (Take the coefficient of static friction between the block and the trolley surface to be 0,8). (A) Zero (B) 5 N (C) 4 N (D) 25 N

Tardigrade Admin · Accepted Answer

Applied force =m × a=5 × 5=25 Limiting friction =μ m g=0.8 × 5 × 10=40 For the given condition applied force is very smaller than limiting friction ∴ Static friction on a body = applied force =m a =5 × 5=25 N

Q. An iron block of mass 5kg is kept on a trolley. If the trolley is being pushed with an acceleration of 5m/s2, what will be the force of friction between the block and the trolley surface? (Take the coefficient of static friction between the block and the trolley surface to be 0,8).

Zero

5 N

4 N

25 N

Applied force =m×a=5×5=25 Limiting friction =μmg=0.8×5×10=40 For the given condition applied force is very smaller than limiting friction ∴ Static friction on a body = applied force =ma =5×5=25N

Applied force $= m \times a = 5 \times 5 = 25$
Limiting friction $= μ m g = 0.8 \times 5 \times 10 = 40$
For the given condition applied force is very smaller than limiting friction
$∴$ Static friction on a body $=$ applied force $= ma$
$= 5 \times 5 = 25 N$