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3 years ago

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A 3.0 kg box is sliding along a frictionless horizontal surface with a speed of 1.8 m/s when it encounters a spring. (a) Determi

ne the force constant (in N/m) of the spring, if the box compresses the spring 6.6 cm before coming to rest.
(b) Determine the initial speed (in m/s) the box would need in order to compress the spring by 1.8 cm.

2 answers:

krok68 [10]3 years ago

8 0

Answer:

a) k = 2231.40 N/m

b) v = 0.491 m/s

Explanation:

Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.

when the box encounters the spring, all the energy of the box is kinetic energy:

the energy relationship between the box and the spring is given by:

1/2(m)×(v^2) = 1/2(k)×(x^2)

(m)×(v^2) = (k)×(x^2)

a) (m)×(v^2) = (k)×(x^2)

k = [(m)×(v^2)]/(x^2)

k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)

k = 2231.40 N/m

Therefore, the force spring constant is 2231.40 N/m

b) (m)×(v^2) = (k)×(x^2)

v^2 = [(k)(x^2)]/m

v = \sqrt{ [(k)(x^2)]/m}

v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}

= 0.491 m/s

bagirrra123 [75]3 years ago

6 0

Answer:

(a) Spring constant k = 2231.4 N/m

(b) Velocity v = 0.49 m/sec

Explanation:

We have given mass of the box m = 3 kg

Speed v = 1.8 m/sec

(a) The spring is stretched by 6.6 cm so x = 6.6 cm = 0.066 m

From energy conservation we know that

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

$\frac{1}{2}\times 3\times 1.8^2=\frac{1}{2}\times k\times 0.066^2$

$k=2231.40N/m$

(b) Here the spring is stretched by 1.8 cm

So x = 1.8 cm = 0.018 m

From energy conservation we know that

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

$\frac{1}{2}\times 3\times v^2=\frac{1}{2}\times 2231.40\times 0.018^2$

$v=0.49m/sec$

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