Question

A horizontal force of 129.4 N is applied on a 10 kg block which rests on a horizontal surface. If the coefficient of friction is 0.3, the acceleration should be (take g = 9.8m * s ^ - 2​

Answer 1

A horizontal force of 129.4 N is applied on a 10 kg block which rests on a horizontal surface.

Coefficient of friction = 0.3 .

First, let's find out the the max static frictional force:

$f_{limiting} = \mu_{s}(N)$

$\implies f_{limiting} = 0.3(10 \times 9.8)$

$\implies f_{limiting} = 29.4 \: N$

Now, net acceleration will be :

$\implies F_{net} = F_{applied} - f_{limiting}$

$\implies F_{net} =129.4- 29.4$

$\implies F_{net} =100$

$\implies a_{net} = \dfrac{100}{m}$

$\implies a_{net} = \dfrac{100}{10}$

$\implies a_{net} = 10 \: m {s}^{ - 2}$

So, net acceleration is 10 m/s².