A force of (200 N) is applied across a (30 Kg) box, calculate the friction force and the vertical force at (a= 2 m/s2). If the force inclined by 60 degree from the horizontal plane (no move in y axis) *

1.) Frictional force = 40N

2.) Vertical force = 294 N

Explanation:

Given that a force of (200 N) is applied across a (30 Kg) box, calculate the friction force and the vertical force at (a= 2 m/s2). If the force inclined by 60 degree from the horizontal plane 

The given parameters are:

Force F = 200 N

Mass M = 30 kg

Acceleration a = 2 m/s^2

Angle Ø = 60 degree

For the object to move horizontally, the force applied must overcome the frictional force.

But force F = product of mass and acceleration

That is,

FcosØ - fr = ma

Where fr = frictional force.

Substitute all the parameters into the formula

200 cos 60 - fr = 30 × 2

100 - fr = 60

-fr = 60 - 100

Fr = 40 N

Therefore, the frictional force is equal to 40N

The vertical force ( normal force) will be equal to the weight of the body

Weight = mg

Weight = 30 × 9.8

Weight = 294 N

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.