Question

33. A block of mass 5 kg is placed on a horizontal
surface with coefficient of friction u = 0.2, then the
maximum and minimum value of force F for which
the block remains at rest are (g = 10 m/s2)
15 N
E
5 kg
(1) 15 N, 10 N
(2) 25 N, 5 N
(4) 5 N, 25 N
(3) 10 N, 25 N​

Answer 1

Answer:

The maximum and minimum value of force are 25 N and 5 N.

(2) is correct option.

Explanation:

Given that,

Mass of block = 5 kg

Coefficient of friction = 0.2

Force = 15 N

We need to calculate the frictional force

Using formula of frictional force

$f_{\mu}=\mu N$

$f_{\mu}=0.2\times5\times10$

$f_{\mu}=10\ N$

We need to calculate the maximum force

Using equation of balance

$F+f_{\mu}=f_{max}$

Put the value into the formula

$f_{max}=15+10$

$f_{max}=25\ N$

We need to calculate the minimum force

Using equation of balance

$f_{min}+f_{\mu}=F$

Put the value into the formula

$f_{min}+10=15$

$f_{min}=15-10$

$f_{min}=5\ N$

Hence, The maximum and minimum value of force are 25 N and 5 N.