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

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

Mass: m = 0.300 kg

Initial compression in spring: x_{1} = 0.160 m

Place datum for gravitational potential energy at position 1.

Position 1: \nu_{1}=0 \quad T_{1}=\frac{1}{2} m ν_{1}^{2}=0

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}

Position 2: T_{2}=\frac{1}{2} m \nu_{A}^{2}=\frac{1}{2}(0.3) \nu_{A}^{2}=0.15 \nu_{A}^{2}

T_{1}+V_{1}=T_{2}+V_{2}: \quad 0+7.68=0.15 \nu_{A}^{2}+2.3544

\nu_{A}^{2}=35.504 \mathrm{~m}^{2} / \mathrm{s}^{2}

Position 3: \quad T_{3}=\frac{1}{2} m \nu_{B}^{2}=\frac{1}{2}(0.3) \nu_{B}^{2}=0.15 \nu_{B}^{2}

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

T_{1}+V_{1}=T_{3}+V_{3}: \quad 0+7.68=0.15 \nu_{B}^{2}+4.7088

\nu_{B}^{2}=19.808 \mathrm{~m}^{2} / \mathrm{s}^{2}

Position 4: T_{2}=\frac{1}{2} m \nu_{C}^{2}=\frac{1}{2}(0.3) \nu_{C}^{2}=0.15 \nu_{C}^{2}

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

T_{1}+V_{1}=T_{4}+V_{4}: \quad 0+7.68=0.15 \nu_{C}^{2}+2.3544

\nu_{C}^{2}=35.504 \mathrm{~m}^{2} / \mathrm{s}^{2}

(a) Newton’s second law at A:

\begin{aligned}& a_{n}= \frac{\nu_{A}^{2}}{\rho}=\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}\right)\end{aligned}

\mathbf{N}_{A}=13.31 \mathrm{~N} \longrightarrow \blacktriangleleft

(b) Newton’s second law at B:

(c) Newton’s second law at C:

\begin{gathered}a_{n}=\frac{\nu_{C}^{2}}{\rho}=\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}\right)\end{gathered}

\mathbf{N}_{C}=13.31 \mathrm{~N}\longleftarrow\blacktriangleleft