A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F → 1 is the force…

A block of mass 5 kg is placed on a rough inclined surface as shown in the figure. If F1 is the force required to just move the block up the inclined plane and F2 is the force required to just prevent the block from sliding down, then the value of F1F2 is: [Use g=10 m s-2]

  1. 253 N
  2. 53 N
  3. 532 N
  4. 10 N

Solution

Limiting friction force is, fmax=μmg cosθ

=0.1×50×32

=2.53 N

For first case:

We can write, F1=mg sinθ+fmax

=25+2.53

For second case:

F2=mg sinθfmax

=252.53

F1F2=53 N

Asked in: JEE Main 2024 (31 Jan Shift 2)

Practice more Laws of Motion questions on Aicharya