GOAL Use conservation of energy to calculate the speed of a block on a horizontal spring with and without friction. PROBLEM A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k = 4.00 x 10? N/m, as in the figure. The surface the block rests upon is frictionless. If the block is pulled out to x, = 0.0500 m and released, (a) ww- find the speed of the block when it first reaches the equilibrium point, (b) find the speed when x= 0.0250 m, and (c) repeat part (a) if friction acts on the block, with coefficient mig - 0.150. A mass attached to a spring. STRATEGY In parts (a) and (b) there are no nonconservative forces, so conservation of energy, can be applied. In part (c) the definition of work and the work-energy theorem are needed to deal with the loss of mechanical energy due to friction. SOLUTION (A) Find the speed of the block at the equilibrium point. Start with the conservation of energy (KE + PE, + PE), = (KE + PE, + PE,), equation. (1) mv? + kn? - mv? + ? Substitute expressions for the block's kinetic energy and the potential energy, and set the gravity terms to zero. Substitute v, - 0, x,- 0, and multiply 자. m by 2/m. Solve for v, and substitute the given 4.00 x 102 N/m (0.0500 m) 5.00 kg X - V values. 0.447 m/s (B) Find the speed of the block at the halfway point. Set v, = 0 in Equation (1) and multiply kx - v+ kx by 2/m. m m Solve for v, and substitute the given V,-V(x? - x) values. 4.00 x 102 N/m(0.050 m) - (0.025 m)?) -V 5.00 kg - 0.387 m/s (C) Repeat part (a), this time with friction. Apply the work-energy theorem. The work done by the force of gravity and the normal force is zero because these forces are perpendicular to the motion. Substitute v, - 0, x, - 0, and Werie -HN,- Set n- mg and solve for mv? -x? - mmgx, v,. v, -Vx? - 2gx, m 4.00 × 10² N/m.(0.050 m)² – 2(0.150)(9.80 m/s*)(0.050 m) 5.00 kg 0.230 m/s LEARN MORE REMARKS Friction or drag from immersion in a fluid damps the motion of an object attached to a spring, eventually bringing the object to rest. QUESTION In the case of friction, what percent of the mechanical energy was lost by the time the mass first reached the equilibrium point? (Hint: use the answers to parts (a) and (c).)

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EXAMPLE 5.9
A Horizontal Spring
GOAL Use conservation of energy to calculate the speed of a
block on a horizontal spring with and without friction.
PROBLEM A block with mass of 5.00 kg is attached to a
horizontal spring with spring constant k = 4.00 x 102 N/m, as
in the figure. The surface the block rests upon is frictionless. If
the block is pulled out to x, = 0.0500 m and released, (a)
x3B
find the speed of the block when it first reaches the equilibrium
point, (b) find the speed when x = 0.0250 m, and (c)
mg
repeat part (a) if friction acts on the block, with coefficient u
= 0.150.
A mass attached to a spring
STRATEGY In parts (a) and (b) there are no nonconservative forces, so conservation of energy, can be
applied. In part (c) the definition of work and the work-energy theorem are needed to deal with the loss of
mechanical energy due to friction.
SOLUTION
(A) Find the speed of the block at the equilibrium point.
Start with the conservation of energy
(KE + PE, + PE,), = (KE + PE, + PE,),
equation.
(1) mv? + ? = mv?
Substitute expressions for the block's
+
kinetic energy and the potential
energy, and set the gravity terms to
zero.
Substitute v, = 0, 1, = 0, and multiply
kx? = v?
m
by 2/m.
Solve for v, and substitute the given
-VE-
k,
4.00 x 102 N/m(0.0500 m)
5.00 kg
V -
values.
= 0.447 m/s
(B) Find the speed of the block at the halfway point.
kx?
= v? +
kx,
Set v, = 0 in Equation (1) and multiply
by 2/m.
m
m
Solve for v, and substitute the given
values.
4.00 x 102 N/m(0.050 m)? - (0.025 m))
5.00 kg
= 0.387 m/s
(C) Repeat part (a), this time with friction.
Apply the work-energy theorem. The
work done by the force of gravity and
Wne = mv? - mv? + ex? - kx?
the normal force is zero because these
forces are perpendicular to the motion.
Substitute v, = 0, x, = 0, and
Wtric = -Hnx,.
Set n- mg and solve for V
Imv? -kx? - Humgx,
.
v, - Vx? - 2gx;
4.00 × 10² N/m0.050 m)? - 2(0.150)(9.80 m/s?)(0.050 m)
5.00 kg
- 0.230 m/s
LEARN MORE
REMARKS Friction or drag from immersion in a fluid damps the motion of an object attached to a
spring, eventually bringing the object to rest.
QUESTION In the case of friction, what percent of the mechanical energy was lost by the time the mass
first reached the equilibrium point? (Hint: use the answers to parts (a) and (c).)
**** .
Transcribed Image Text:EXAMPLE 5.9 A Horizontal Spring GOAL Use conservation of energy to calculate the speed of a block on a horizontal spring with and without friction. PROBLEM A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k = 4.00 x 102 N/m, as in the figure. The surface the block rests upon is frictionless. If the block is pulled out to x, = 0.0500 m and released, (a) x3B find the speed of the block when it first reaches the equilibrium point, (b) find the speed when x = 0.0250 m, and (c) mg repeat part (a) if friction acts on the block, with coefficient u = 0.150. A mass attached to a spring STRATEGY In parts (a) and (b) there are no nonconservative forces, so conservation of energy, can be applied. In part (c) the definition of work and the work-energy theorem are needed to deal with the loss of mechanical energy due to friction. SOLUTION (A) Find the speed of the block at the equilibrium point. Start with the conservation of energy (KE + PE, + PE,), = (KE + PE, + PE,), equation. (1) mv? + ? = mv? Substitute expressions for the block's + kinetic energy and the potential energy, and set the gravity terms to zero. Substitute v, = 0, 1, = 0, and multiply kx? = v? m by 2/m. Solve for v, and substitute the given -VE- k, 4.00 x 102 N/m(0.0500 m) 5.00 kg V - values. = 0.447 m/s (B) Find the speed of the block at the halfway point. kx? = v? + kx, Set v, = 0 in Equation (1) and multiply by 2/m. m m Solve for v, and substitute the given values. 4.00 x 102 N/m(0.050 m)? - (0.025 m)) 5.00 kg = 0.387 m/s (C) Repeat part (a), this time with friction. Apply the work-energy theorem. The work done by the force of gravity and Wne = mv? - mv? + ex? - kx? the normal force is zero because these forces are perpendicular to the motion. Substitute v, = 0, x, = 0, and Wtric = -Hnx,. Set n- mg and solve for V Imv? -kx? - Humgx, . v, - Vx? - 2gx; 4.00 × 10² N/m0.050 m)? - 2(0.150)(9.80 m/s?)(0.050 m) 5.00 kg - 0.230 m/s LEARN MORE REMARKS Friction or drag from immersion in a fluid damps the motion of an object attached to a spring, eventually bringing the object to rest. QUESTION In the case of friction, what percent of the mechanical energy was lost by the time the mass first reached the equilibrium point? (Hint: use the answers to parts (a) and (c).) **** .
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