Question

A block of wood weights 10 N and is resting on an inclined plane. The coefficient of friction is 0.7. The frictional force that acts on the block when the plane is ${30}^o$ inclined with the horizontal is:

• A. 6.062 N
• B. 5 N
• C. 9.8 N
• D. 70 N

Answer 1

The correct option is A 6.062 N

Given that,

Weight of block = 10 N

Mass = $\frac{10}{10}=1$ kg

Coefficient of friction $\mu =0.7$

We know that,

Normal reaction = mg cosθ

$N=mg\cos \theta$

$N=W\cos \theta$

$N=10\times \cos {30}^0$

$N=10\times \frac{\sqrt{3}}{2}$

$N=5\sqrt{3}$

Now, frictional force

$F=0.7\times 5\sqrt{3}$

$F=6.062N$

Hence, the frictional force is 6.062 N.