Question

A horizontal spring with spring constant 290 N/m is compressed by 10 cm and then used to launch a 300 g box across the floor. The coefficient of kinetic friction between the box and the floor is 0.23.What is the box's launch speed?

Answer 1

Answer:

3.03568 m/s

Explanation:

$\mu$ = Coefficient of kinetic friction = 0.23

k = Spring constant = 290 N/m

x = Compression distance = 0.1 m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of box = 0.3 kg

$F_n$ = Normal force = $mg$

v = Launch velocity

Potential energy of spring

$U=\frac{1}{2}kx^2\\\Rightarrow U=\frac{1}{2}\times 290\times 0.1^2\\\Rightarrow U=1.45\ J$

As the potential and kinetic energies are conserved

$U=K+W_f\\\Rightarrow U=\frac{1}{2}mv^2+\mu F_nd\\\Rightarrow U=\frac{1}{2}mv^2+\mu mgx\\\Rightarrow v=\sqrt{2\frac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\frac{1.45-0.23\times 0.3\times 9.81\times 0.1}{0.3}}\\\Rightarrow v=3.03568\ m/s$

The launch speed of the box is 3.03568 m/s

Answer 2

The launch speed of the box across the floor is 1.52 m/s.

The given parameters;

• elastic constant of the spring, k = 290 N/m
• extension of the spring, x = 10 cm
• mass of the box, m = 300 g = 0.3 kg
• coefficient of kinetic friction, μ = 0.23

Apply the principle of conservation of energy to determine the launch speed of the box;

$\frac{1}{2} kx^2 - \mu mgx = \frac{1}{2} mv_i^2\\\\(0.5\times 290\times 0.1^2 )-(0.23 \times 0.3\times 9.8\times 0.1) = \frac{1}{2} \times 0.3\times v_i^2\\\\1.382 = 0.6v_i^2\\\\v_i^2 = \frac{1.382}{0.6} \\\\v_i^2 = 2.303\\\\v_i = \sqrt{2.303} \\\\v_i = 1.52 \ m/s$

Thus, the launch speed of the box across the floor is 1.52 m/s.

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