7. A 2kg block is acted on by two forces F1 and F2 as shown in the diagram. If the magnitude of F1 = 13N and F2 = 11N. If the coefficient of kinetic friction is .23, find (A) The normal force exerted on it by the floor. (2 pts) (B) The acceleration on of the block (2 pts)

A.) 17.04 N

B.) 10.76 m/s^2

Explanation:

Given that a 2kg block is acted on by two forces F1 and F2 as shown in the diagram. If the magnitude of F1 = 13N and F2 = 11N. If the coefficient of kinetic friction is .23,

A.) To find the normal force exerted on it by the floor, we need to first calculate the vertical forces of the ropes.

The vertical force on F1 = 13 sin 30

F1 = -6.5 N

The vertical force on F2 = 11 sin 21

F2 = 3.94

The resultant force = - 6.5 + 3.94

Resultant force = - 2.56

The normal force = mg - 2.56

The normal force = 2 × 9.8 - 2.56

The normal force = 19.6 - 2.56

The normal force = 17.04 N

(B) The acceleration on of the block will be achieved by first calculating the horizontal force.

For F1 = 13Cos 30

F1 = 11.26N

For F2 = 11 cos 21

F2 = 10.27

the resultant force = 11.26 + 10.27

Horizontal force = 21.53N

Force = mass × acceleration

21.53 = 2 × a

a = 21.53 / 2

Acceleration a = 10.76 m/s^2

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards