A crate of mass m = 0.25 kg is set against a spring with spring constant of k_1

Question

A crate of mass m = 0.25 kg is set against a spring with spring constant of k_1 = 613N/m which has been compressed by a distance of 0 1m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k_2 = 452N/m. How far, d_2 in meters, will the second spring compress when the crate runs into it? How fast, v in meters per second, will the crate be moving when it strikes the second spring? Now assume that the surface is rough. You perform the experiment and you observe that second spring only compresses a distance of d_2/2. How much energy, in joules, was lost to friction?

Explanation / Answer

here,

mass of crate , m = 0.25 kg

k1 = 613 N/m

compression in spring 1 , x1 = 0.1 m

k2 = 452 N/m

(a)

let the compression in the seccond spring be d2

using conervation of energy

0.5 * k1 * x1^2 = 0.5 * k2 * d2^2

613 * 0.1^2 = 452 * d2^2

d2 = 0.12 m

the compression in the seccond spring is 0.12 m

(b)

let the speed be v

using conservation of energy

energy stored in spring 1 = kinetic energy of crate

0.5 * k1 * x1^2 = 0.5 * m * v^2

613 * 0.1^2 = 0.25 * v^2

v = 4.95 m/s

the speed of crate before it strikes seccond spring is 4.95 m/s

(c)

compression in spring 2 , d = d2/2 = 0.06 m

the energy lost due to friction , E = energy stored in spring 1 - energy stored in spring 2

E = 0.5 * ( k1 * x1^2 - k2 * d^2)

E = 0.5 * ( 613 * 0.1^2 - 452 * 0.06^2)

E = 2.25 J

the energy lost due to friction is 2.25 J

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