Question

A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0 cm before coming to rest. Determine the force constant (in N/m) of the spring, if the coefficient of kinetic friction between the block and the surface is μk = 0.660.

Answer 1

Answer:

463.2 N/m

Explanation:

mass of block, m = 2 kg

velocity of spring, v = 1.9 m/s

distance, r = 10 cm = 0.1 m

coefficient of friction, μ = 0.66

Let the spring constant is K.

friction force, f = μ mg = 0.66 x 2 x 9.8 = 12.94 N

Use work energy theorem

$\frac{1}{2}mv^{2}=f \times r + \frac{1}{2}Kr^{2}$

0.5 x 2 x 1.9 x 1.9 = 12.94 x 0.1 + 0.5 x K x 0.1 x 0.1

3.61 = 1.294 + 0.005 K

K = 463.2 N/m

Thus, the spring constant is 463.2 N/m.

Answer 2

Answer:

Explanation:

Given

mass of box $m=2\ kg$

speed of box $v=1.9\ m/s$

distance moved by the box $x=10\ cm$

coefficient of kinetic friction $\mu _k=0.66$

Friction  force $f_r=\mu_kN$

$f_r=0.66\times mg$

$f_r=0.66\times 2\times 9.8=12.936 \N$

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

$\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2$

$\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2$

$3.61-1.2936=0.005\times k$

$k=463.28\ N/m$