Question

A block with mass m = 0.250 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at x = +0.240 m, its acceleration is ax = -13.0 m/s2 and its velocity is vx = +4.00 m/s.1. What is the spring's force constant k?2. What is the amplitude of the motion?3. What is the maximum speed of the block during its motion?4. What is the maximum magnitude of the block's acceleration during its motion?

Answer 1

Answer:

13.54 N/m

0.6 m

4.37 m/s

32.496 m/s²

Explanation:

m = Mass of block = 0.25 kg

k = Spring constant

A = Amplitude

x = Compression of spring = 0.24 m

a = Acceleration = -13 m/s²

v = Velocity = 4 m/s

The weight of the block and force on spring is equal

$ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m$

The spring's force constant is 13.54 N/m

Total energy of the system is given by

$E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J$

At maximum displacement v = 0

$E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)$

$\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m$

The amplitude of the motion is 0.6 m

Speed of the block

$E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s$

The maximum speed of the block during its motion is 4.37 m/s

Forces in the spring

$ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2$

Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²