Respuesta :
Answer:
The acceleration of the block is 6.35 m/s².
Explanation:
It is given that,
A force is applied to a block to move it up a 30° incline. The applied force is, F = 65 N
Mass of the block, m = 5 kg
We need to find the acceleration of the block. From the attached figure, it is clear that.
[tex]F_x=ma_x[/tex]
[tex]F\ cos\theta-mg\ sin\theta=ma_x[/tex]
[tex]a_x=\dfrac{F\ cos\theta-mg\ sin\theta}{m}[/tex]
[tex]a_x=\dfrac{65\ N\ cos(30)-5\ kg\times 9.8\ m/s^2\ sin(30)}{5\ kg}[/tex]
[tex]a_x=6.35\ m/s^2[/tex]
So, the acceleration of the block is 6.35 m/s². Hence, this is the required solution.
Acceleration is defined as the rate of change of the velocity of the body. Its unit is m/sec².The magnitude of the acceleration of the block will be 6.35 m/sec².
What is the friction force?
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.
On resolving the given force and accelertaion in the different components and balancing the equation gets.Components in the x-direction.
The given data in the problem,
F is the applied force =65 N
Θ is the angle of inclined plane=30°
m is the mass of the block= 5 kg
We need to find the acceleration of the block in the x-direction
[tex]\rm F_x=ma_x\\\\\rm F_x=Fcos\theta-mgsin\theta\\\\\rm Fcos\theta-mgsin\theta=ma_x\\\\\rm a_x=\frac{Fcos\theta-mgsin\theta}{m}\\\\ \rm a_x=\frac{65\timescos30^0-mgsin30^0}{5} \\\\\rm a_x=6.35 m/sec^2[/tex]
Hence the magnitude of the acceleration of the block will be 6.35 m/sec².
To know more about friction force refer to the link;
https://brainly.com/question/1714663
