
Concept explainers
A 10-kg block is attached to spring A and connected to spring B by a cord and pulley. The block is held in the position shown with both springs unstretched when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 2 kN/m, determine (a) the velocity of the block after it has moved down 50 mm, (b) the maximum velocity achieved by the block.
Fig. P13.30
(a)

Find the velocity (v) of the block after it has moved down 50mm.
Answer to Problem 13.30P
The velocity (v) of the block after it has moved down 50mm is 0.597m/s_.
Explanation of Solution
Given information:
The mass of the block (m) is 10kg.
The spring constant at A (kA) is 2kN/m.
The spring constant at B (kB) is 2kN/m.
The depth where the spring A moves down (xA) is 50mm or 0.05m.
Assume the acceleration due to gravity (g) is 9.81m/s2.
Calculation:
Show the free body diagram of the block with two spring’s attachment acting as in Figure (1).
Calculate the depth of spring B moves down due to block (xB) using the formula:
xB=12xA
Substitute 0.05m for xA.
xB=12(0.05)=0.025m
Calculate the weight of the block (W) using the relation:
W=mg
Substitute 10kg for m and 9.81m/s2 for g.
W=(10)(9.81)=98.1N
Here, the initial kinetic energy (T1) is zero respectively.
Calculate the final kinetic energy (T2) using the formula:
T2=12mv2
Substitute 10kg for m.
T2=12(10)v2=5v2
Calculate the work done (U1−2)G due to gravity using the formula:
(U1−2)G=WxA
Substitute 98.1N for W and 0.05m for xA.
(U1−2)G=(98.1)0.05=4.905N⋅m
Calculate the work done (U1−2)A due to spring A using the formula:
(U1−2)A=12kAx2A
Substitute 0.05m for xA and 2kN/m for kA.
(U1−2)A=12(2×103)(0.05)2=2.5N⋅m
Calculate the work done (U1−2)B due to spring B using the formula:
(U1−2)B=12kBx2B
Substitute 0.025m for xB and 2kN/m for kB.
(U1−2)B=12(2×103)(0.025)2=0.625N⋅m
Calculate the total work done (U1−2) using the relation:
(U1−2)=(U1−2)G−(U1−2)A−(U1−2)B
Substitute 4.905N⋅m for (U1−2)G, 2.5N⋅m for (U1−2)A, and 0.625N⋅m for (U1−2)B.
(U1−2)=4.905−2.5−0.625=1.78N⋅m
Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.
Find the velocity (v) of the block after it has moved down 50mm:
T1+U1−2=T2
Substitute 0 for T1, 1.78N⋅m for U1−2, and 5v2 for T2.
0+(1.78N⋅m)=5v2v=√1.785v=0.597m/s
Therefore, the velocity (v) of the block after it has moved down 50mm is 0.597m/s_.
(b)

Find the maximum velocity (vmax) achieved by the block.
Answer to Problem 13.30P
The maximum velocity (vmax) achieved by the block is 0.620m/s_
Explanation of Solution
Given information:
The mass of the block (m) is 10kg.
The spring constant at A (kA) is 2kN/m.
The spring constant at B (kB) is 2kN/m.
The depth where the spring A moves down (xA) is 50mm or 0.05m.
Assume the acceleration due to gravity (g) is 9.81m/s2.
Calculation:
Assume x be the distance moved down by the 10kg block.
Calculate the work done (U1−2)G due to gravity using the formula:
(U1−2)G=Wx
Substitute 98.1N for W.
(U1−2)G=98.1x
Calculate the work done (U1−2)A due to spring A using the formula:
(U1−2)A=12kAx2
Substitute 2kN/m for kA.
(U1−2)A=12(2×103)x2=1000x2
Calculate the work done (U1−2)B due to spring B using the formula:
(U1−2)B=12kB(x2)2
Substitute 2kN/m for kB.
(U1−2)B=12(2×103)(x2)2=1000(x24)=250x2
Calculate the total work done (U1−2) using the relation:
(U1−2)=(U1−2)G−(U1−2)A−(U1−2)B
Substitute 98.1x for (U1−2)G, 1000x2 for (U1−2)A, and 250x2 for (U1−2)B.
(U1−2)=98.1x−1000x2−250x2=98.1x−1250x2 (1)
Differentiate the above equation with respect to ‘x’.
dU1−2dx=0
Substitute 98.1x−1250x2 for U1−2.
ddx(98.1x−1250x2)=098.1−2500x=0x=98.12,500x=0.03924m(39.24mm)
Substitute 0.03924m for x in Equation (1).
(U1−2)=98.1(0.03924)−1000(0.03924)2−250(0.03924)2=1.925N⋅m
Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.
Find the maximum velocity (vmax) achieved by the block:
T1+U1−2=T2
Substitute 0 for T1, 1.925N⋅m for U1−2, and 5v2 for T2.
0+(1.925N⋅m)=5v2v=√1.9255v=0.620m/s
Therefore, the maximum velocity (vmax) achieved by the block is 0.620m/s_.
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Chapter 13 Solutions
Vector Mechanics for Engineers: Statics and Dynamics
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