A block having a mass of 0.70 kg is given an initial velocity VA 1.4 m/s to the
ID: 1584281 • Letter: A
Question
A block having a mass of 0.70 kg is given an initial velocity VA 1.4 m/s to the right and collides with a spring whose mass is negligible and whose force constant is k = 51 N/m as shown in the figure (A) Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision (B) Suppose a constant force of kinetic friction acts between the block and the surface, with .-0.51. If the speed of the block at the moment it collides with the spring is VA1.4 m/s, what is the maximum compression xc in the spring? A block sliding on a frictionless, horizontal surface collides with a light spring. (a) Initially, the mechanical energy is all kinetic energy. (b) The mechanical energy is the sum of the kinetic energy of the block and the elastic potential energy in the spring (c) The energy is entirely potential energy. (d) The energy is transformed back to the kinetic energy throughout the motion. of the block. The total energy of the system remains constantExplanation / Answer
(A)
Just before collision,
Potential energy of block, UAg = 0 J (gravitational potential energy)
Kinetic energy of block, KA = 0.5mvA2= (0.5)(0.70)(1.4)2 = 0.686 J
Total energy of block, EA = 0 + 0.686 = 0.686 J
After collision when block-spring system comes to rest,
Gravitational potential energy of block-spring system, UCg = 0 J
Spring potential energy of block-spring system, UCs = 0.5k(xmax)2
Kinetic energy of block-spring system, KC = 0 J
Total energy of block-spring system, EC = 0 + 0.5k(xmax)2 + 0 = 0.5k(xmax)2
Using energy conservation,
EA = EC
0.686 = 0.5k(xmax)2
(xmax)2 = 0.686/(0.5k)
(xmax)2 = 0.686/(0.5*51)
xmax = 0.164 m
(B)
Just before collision,
Potential energy of block, UAg = 0 J (gravitational potential energy)
Kinetic energy of block, KA = 0.5mvA2= (0.5)(0.70)(1.4)2 = 0.686 J
Total energy of block, EA = 0 + 0.686 = 0.686 J
After collision during the motion of block-spring system to come to rest,
Amount of work done, W = -fx = -(µmg)(xc) = -(0.51*0.70*9.8)(xc) = -3.4986xc
After collision when block-spring system comes to rest,
Gravitational potential energy of block-spring system, UCg = 0 J
Spring potential energy of block-spring system, UCs = 0.5k(xmax)2
Kinetic energy of block-spring system, KC = 0 J
Total energy of block-spring system, EC = 0 + 0.5k(xmax)2 + 0 = 0.5k(xmax)2
Using energy conservation,
EA= EC-3.4986xc
0.686 = 0.5k(xc)2 -3.4986xc
0.5k(xc)2 -3.4986xc – 0.686= 0
25.5(xc)2 -3.4986xc – 0.686= 0
xc = 0.246 m.
Master it
If we assume the top of compressed spring as the reference level for gravitational potential energy, then
Just before collision,
Potential energy of block, UAg = mgxmax = 0.7*9.8*xmax = 6.86xmax (gravitational potential energy)
Kinetic energy of block, KA = 0.5mvA2= (0.5)(0.70)(1.8)2 = 1.134 J
Total energy of block, EA = 6.86xmax + 1.134
After collision when block-spring system comes to rest,
Gravitational potential energy of block-spring system, UCg = 0 J
Spring potential energy of block-spring system, UCs = 0.5k(xmax)2
Kinetic energy of block-spring system, KC = 0 J
Total energy of block-spring system, EC = 0 + 0.5k(xmax)2 + 0 = 0.5k(xmax)2
Using energy conservation,
EA = EC
6.86xmax + 1.134 = 0.5k(xmax)2
0.5k(xmax)2 - 6.86xmax - 1.134 = 0
25.5k(xmax)2 - 6.86xmax - 1.134 = 0
xmax = 0.385 m
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