Question

A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m . When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The spring constant k is 100 N/m. What is the coefficient of kinetic friction between the block and the tabletop

Answer 1

Answer:

μ = 0.41

Explanation:

$\frac{1}{2} kx^2 = \mu mgd$

$\mu = (\frac{1}{2} kx^2)/mgd$

$\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)$

$\mu = 0.41$

Answer 2

The coefficient of kinetic friction between the block and the tabletop is 0.41.

The given parameters;

• mass of the block, m = 0.5 kg
• extension of the spring, x = 0.2 m
• spring constant, K = 100 N/m

The coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.

Fd = Uₓ

μmgd = ¹/₂kx²

where;

• μ is the coefficient of kinetic friction

The coefficient of kinetic friction between the block and the tabletop;

$\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41$

Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.

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