Answer:
The acceleration is 5.66m/s²
The frictional force is 11.33N
Explanation:
Please see attached a rough sketch of the free body diagram for your reference
Given data
Mass m=3kg
Angle =39°
We know that the balancing force is given as
mgsinθ-f=ma
And the balancing torque is given as
FR=2/3mR²∝
Please see the attached for the annotated solution
(a) The acceleration of the block will be 5.6 m/sec².
(b)The value of friction force will be 11.33 N.
It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically it is defined as the product of the coefficient of friction and normal reaction.
a) The acceleration of the block will be 5.6 m/sec²
The balancing force is obtained by the formula
[tex]\rm mgsin\theta- F=ma \\\\ f=\frac{2}{3} ma\\\\ \rm a=\frac{3g sin\theta}{5}\\\\ \rm a=\frac{3 \times 9.81 sin39^0}{5} \\\\ \rm a=\frac{28.37}{5} \\\\ \rm a=5.66 m/sec^2[/tex]
Hence the acceleration of the block will be 5.6 m/sec².
(b)The value of friction force will be 11.33 N.
[tex]\rm F=\frac{2}{3}ma\\\\ \rm F=\frac{2}{3}\times 3\times 5.66\\\\ \rm F=11.33 N[/tex]
Hence the value of friction force will be 11.33 N.
To know more about friction force refer to the link;
https://brainly.com/question/1714663
