Question

A 14.7 kg object sits at rest on a 4.0 m long table. One half of the table is smooth (frictionless) and the other half is rough. A constant horizontal force of 12.3 N is applied to the object. (a) What is the speed of the object after it has traveled 2.0 m? At 2.0 m, the constant force is removed and the block comes to rest after traveling 1.75 m along the rough surface. (b) What is the magnitude and direction of the force that brought the block to rest?

Answer 1

a) use Work-energy theorem

Workdone = change in kinetic energy

F*d*cos(0) = (1/2)*m*v^2 - (1/2)*m*vo^2

F*d = (1/2)*m*v^2 (since vo = 0 )

==> v = sqrt(2*F*d/m)

= sqrt(2*12.3*2/14.7)

= 1.83 m/s <<<<<<<<<----------------Answer

b) let fk is the frictional force that stops the body.

again use, work-energy theorem

Workdone by friction = change in kinetic energy

fk*x*cos(180) = (1/2)*m*vf^2 - (1/2)*m*v^2

-fk*x = -(1/2)*m*v^2 ( since vf = 0 )

fk = m*v^2/(2*x)

= 14.7*1.83^2/(2*1.75)

= 14.1 N <<<<<<<<<<----------------Answer

direction : opposite to the motion of the object.