A body weighing 980N is moving with constant speed of 5 m/s on a horizontal surface. The coefficient of sliding friction between the body and the surface is 0.25. What horizontal pull is necessary to maintain the motion?
forceFriction=.25*980
M*g = 980N. = Wt. of body = Normal(Fn).
Fk = u*Fn = 0.25 * 980 = 245 N. = Force of kinetic friction.
F-Fk = M*a.
F-245 = M*0 = 0,
F = 245 N.
To determine the horizontal pull necessary to maintain the motion of the body, we need to consider the forces acting on the body.
First, let's calculate the force of friction that opposes the motion of the body. The force of friction (Ff) can be calculated using the formula:
Ff = coefficient of friction * normal force
The normal force (Fn) is the force exerted by the surface on the body vertically upwards. In this case, the weight of the body (980N) provides the magnitude of the vertical force. So, Fn = 980N.
Substituting the values into the formula, we have:
Ff = 0.25 * 980N
Ff = 245N
Since the body is moving with constant speed, the opposing force of friction must be balanced by an equal and opposite horizontal force (Fpull). Therefore, we can determine the required horizontal force (Fpull) by setting Ff equal to Fpull:
Fpull = Ff
Fpull = 245N
Hence, a horizontal pull of 245N is necessary to maintain the motion of the body.