Question

A 100 N force acts horizontally on a block of 10
kg placed on a horizontal rough surface of
coefficient of friction u=0.5. If the acceleration
due to gravity (g) is taken as 10 ms, the
acceleration of the block (in ms-2) is
(a) 2.5
(b) 10
(c) 5
(d) 7.5
solve it.​

Answer 1

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

Refer to attachment:

GIVEN:

→Mass of the block is 10

→A force of 100N is acting on the block.

→Coefficient$\mu_{k}$of friction in 0.5.

TO FIND:

The acceleration of the block.

SOLUTION:

Let the acceleration of the block be 'a'.

WE KNOW,

$\large\red{\boxed{Force_{f}=\mu_{k}N}}$

$\large\red{\boxed{Force_{f}=\mu_{k}N}}$

where, N is normal reaction

& F is frictional force.

Now, here,

$\large\green{\boxed{N =mg}}$

=>$N = (10×10) N$

=>$N = 100N$

______________________________________

=>$F_{f} =\mu_{k}N$

=>$F_{f} = \dfrac{5}{10}\times 100)N$

=>$F_{f}=50N$

______________________________________

So, Net force

$\large\purple{\boxed{F_{net}=F-F_{f}}}$

=>$F_{net}=(100-50) N$

= >$F_{net}=50N$

______________________________________

Now, we know

$\large\blue{\boxed{F_{net}=m × a}}$

=>$50N = 10kg × a$

=>$a =\dfrac{50}{10}m/s^{2}$

=>$a = 5m/s^{2}$

Therefore acceleration of the block is $5m/s^{2}$.

$\huge\orange{\boxed{Accel^{n}=5ms^{-2}}}$