A 300-g block is released from rest after a spring of constant k = 600 N/m has been compressed 160 mm. Determine the force exerted by the loop ABCD on the block as the block passes through (a) Point A, (b) Point B, (c) Point C. Assume no friction.

Question

A 300-g block is released from rest after a spring of constant k = 600 N/m has been compressed 160 mm. Determine the force exerted by the loop ABCD on the block as the block passes through (a) Point A, (b) Point B, (c) Point C. Assume no friction.

Accepted Answer

Conservation of energy to determine speeds at locations A, B, and C.

Mass: m = 0.300 kg

Initial compression in spring: [latex]x_{1}[/latex] = 0.160 m

Place datum for gravitational potential energy at position 1.

Position 1: [latex] u_{1}=0 \quad T_{1}=\frac{1}{2} m ν_{1}^{2}=0[/latex]

[latex]V_{1}=\frac{1}{2} k x_{1}^{2}=\frac{1}{2}(600 \mathrm{~N} / \mathrm{m})(0.160 \mathrm{~m})^{2}=7.68 \mathrm{~J}[/latex]

Position 2: [latex]T_{2}=\frac{1}{2} m u_{A}^{2}=\frac{1}{2}(0.3) u_{A}^{2}=0.15 u_{A}^{2}[/latex]

[latex]T_{1}+V_{1}=T_{2}+V_{2}: \quad 0+7.68=0.15 u_{A}^{2}+2.3544[/latex]

[latex] u_{A}^{2}=35.504 \mathrm{~m}^{2} / \mathrm{s}^{2}[/latex]

Position 3: [latex]\quad T_{3}=\frac{1}{2} m u_{B}^{2}=\frac{1}{2}(0.3) u_{B}^{2}=0.15 u_{B}^{2}[/latex]

[latex]V_{3}=m g h_{3}=(0.3 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2} ight)(1.600 \mathrm{~m})=4.7088 \mathrm{~J}[/latex]

[latex]T_{1}+V_{1}=T_{3}+V_{3}: \quad 0+7.68=0.15 u_{B}^{2}+4.7088[/latex]

[latex] u_{B}^{2}=19.808 \mathrm{~m}^{2} / \mathrm{s}^{2}[/latex]

Position 4: [latex]T_{2}=\frac{1}{2} m u_{C}^{2}=\frac{1}{2}(0.3) u_{C}^{2}=0.15 u_{C}^{2}[/latex]

[latex]V_{4}=m g h_{4}=(0.3 \mathrm{~kg})(9.81 \mathrm{~m} / \mathrm{s})(0.800 \mathrm{~m})=2.3544 \mathrm{~J}[/latex]

[latex]T_{1}+V_{1}=T_{4}+V_{4}: \quad 0+7.68=0.15 u_{C}^{2}+2.3544[/latex]

[latex] u_{C}^{2}=35.504 \mathrm{~m}^{2} / \mathrm{s}^{2}[/latex]

(a) Newton’s second law at A:

[latex]\begin{aligned}& a_{n}= \frac{ u_{A}^{2}}{ ho}=\frac{35.504 \mathrm{~m}^{2} / \mathrm{s}^{2}}{0.800 \mathrm{~m}}=44.38 \mathrm{~m} / \mathrm{s}^{2} \& \mathbf{a}_{n}=44.38 \mathrm{~m} / \mathrm{s}^{2} \longrightarrow \& \overset{+}{\longrightarrow } \Sigma F=m a_{n}: \quad N_{A}=m a_{n} \& N_{A}=(0.3 \mathrm{~kg})\left(44.38 \mathrm{~m} / \mathrm{s}^{2} ight)\end{aligned}[/latex]

[latex]\mathbf{N}_{A}=13.31 \mathrm{~N} \longrightarrow \blacktriangleleft[/latex]

(b) Newton’s second law at B:

[latex]\begin{gathered}a_{n}=\frac{ u_{B}^{2}}{ ho}=\frac{19.808 \mathrm{~m}^{2} / \mathrm{s}^{2}}{0.800 \mathrm{~m}}=24.76 \mathrm{~m} / \mathrm{s}^{2} \\mathbf{a}_{n}=24.76 \mathrm{~m} / \mathrm{s}^{2} \downarrow \+\downarrow \Sigma F=m a_{n}: \quad N_{B}=m g=m a_{n} \N_{B}=m\left(a_{n}-g ight)=(0.3 \mathrm{~kg})\left(24.76 \mathrm{~m} / \mathrm{s}^{2}-9.81 \mathrm{~m} / \mathrm{s}^{2} ight)\end{gathered}[/latex]

[latex]\mathbf{N}_{B}=4.49 \mathrm{~N} \downarrow\blacktriangleleft[/latex]

(c) Newton’s second law at C:

[latex]\begin{gathered}a_{n}=\frac{ u_{C}^{2}}{ ho}=\frac{35.504 \mathrm{~m}^{2} / \mathrm{s}^{2}}{0.800 \mathrm{~m}}=44.38 \mathrm{~m} / \mathrm{s}^{2} \\mathbf{a}_{n}=44.38 \mathrm{~m} / \mathrm{s}^{2} \longleftarrow \\overset{+}{\longleftarrow } \Sigma F=m a_{n}: \quad N_{C}=m a_{n} \N_{C}=(0.3 \mathrm{~kg})\left(44.38 \mathrm{~m} / \mathrm{s}^{2} ight)\end{gathered}[/latex]

[latex]\mathbf{N}_{C}=13.31 \mathrm{~N}\longleftarrow\blacktriangleleft[/latex]

Question 13.195: A 300-g block is released from rest after a spring of consta...

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