Question

A block sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k=432 Nm ; the other end of the spring is fixed in place. The block has a kinetic energy of 20.7 J as it passes through the spring's equilibrium position. As the block slides, a frictional force of magnitude 24 N acts on it. (a) How far will the block slide from the equilibrium position before coming momentarily to rest? (b) What will be the kinetic energy of the block as it slides back through the equilibrium position?

Answer 1

Answer:

a)

$0.26$ m

b)

$8.2$ J

Explanation:

a)

Consider the motion of the block between equilibrium point and the point where it comes to a stop

$KE_{o}$ = initial Kinetic energy of the block at equilibrium position = 20.7 J

$k$ = spring constant of the spring = 432 Nm⁻¹

$f$ = frictional force acting on the block = 24 N

$x$ = stretch in the spring caused before it stops

Using conservation of energy at initial equilibrium point and stopping point

Kinetic energy at equilibrium position = Spring energy at the stopping point + magnitude of work done by frictional force

$KE_{o} = (0.5)kx^{2} + fx$

inserting the values

$20.7 = (0.5)(432)x^{2} + (24)x$

$(216)x^{2} + (24)x - 20.7 = 0$

$x = 0.26$ m

b)

Consider the motion of the block between equilibrium point and the same point when it returns

$KE_{o}$ = initial Kinetic energy of the block at equilibrium position = 20.7 J

$KE_{o}$ = final Kinetic energy of the block at equilibrium position while returning

$f$ = frictional force acting on the block = 24 N

$d$ = distance traveled by the block = $x + x$ =  $2x$

Using conservation of energy at initial equilibrium point and stopping point

Kinetic energy at equilibrium position = Kinetic energy at equilibrium while returning + magnitude of work done by frictional force

$KE_{o} = KE_{o} + 2 fx$

inserting the values

$20.7 = KE_{o} + 2 (24)(0.26)$

$20.7 = KE_{o} + 12.5$

$KE_{o} = 8.2$ J