A 1.10-kg object slides to the right on a surface having a coefficient of kineti
ID: 1490016 • Letter: A
Question
A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of v_i = 3.10 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e). Find the distance of compression d. Find the speed v at the unstretched position when the object is moving to the left (Figure d) Find the distance D where the object comes to rest.Explanation / Answer
a) Use the law of conservation of energy
Kinetic energy of mass = work absorbed by spring + energy to overcome friciton
KE = Ws + Wf
(0.5)(mV2) = (0.5)kd2 + (mu)(m)(g)d
where m = mass = 1.10 kg (given)
V = velocity of mass = 3.10 m/sec. (given)
k = spring constant = 50 N/m (given)
d = distance at which spring was compressed
mu = coefficient of friction = 0.250 (given)
g = acceleration due to gravity = 9.8 m/s2 (constant)
Substituting values,
(0.5)(1.10)(3.10)2 = (0.5)(50)d2 + (0.250)(1.10)(9.8)(d)
5.2855 = 25d2 + 2.695(d)
Rearranging the above,
25d2 + 2.695d - 5.2855= 0
Using the quadratic equation,
d = 0.4091 m.
b) Use the law of conservation of energy again, i.e.,
Energy from spring = Kinetic energy of mass + Work to overcome friction
Ws = KE + Wf
(0.5)(50)(0.4091)2 = (0.5)(1.10)V2 + (0.25)(1.10)(9.8)(0.4091)
4.184 = 0.55(V2) + 1.1025
0.55(V2) = 3.0815
V = 2.37 m/sec.
c) Working formula is
Vf2 - V2 = 2aD
where Vf = final velocity = 0 (when object comes to rest)
V = 2.37 m/sec (as determined above)
a = acceleration
D = distance when mass will stop
From Newton's 2nd Law of Motion, F = ma
where f = frictional force = (mu)(mg)
Therefore, (mu)(mg) = ma
and since "m" appears on both sides of the equation, it will simply cancel out, hence
a = 0.250*9.8 = 2.45 m/s2
Substituting values,
0 - 2.372 = 2(-2.45)(D)
NOTE the negative sign attached to the acceleration. This simply means that the mass was slowing down as it was moving away from the spring.
Solving for "D"
D = 1.15 m.
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