A rightward force is applied to a 10-kg object to move it across a rough surface at constant velocity. The coefficient of friction between the object and the surface is 0.2. Use the diagram to determine the (A) gravitational force, (B) normal force, (C) applied force, (D) frictional force, and (E) net force. (Neglect air resistance.)

Explanation:

Gravitational force = normal force which are both the same thing as the weight of the object which is found in w = mg. Therefore,

w = 10(9.8) = 98 N.

If the object is being pulled at a constant velocity, then the acceleration of said object is 0. Thus,

F - f = ma and

F - f = 0 so

F = f.  So we just need to solve for f, which is found in the equation

f = μ so

f = .2(98) and

f = 19.6 = F

The net force is 0 if F = f

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.