Question

In: Physics

a crate of stuffed kittens with a mass of 1kg compresses a spring a distance of...

a crate of stuffed kittens with a mass of 1kg compresses a spring a distance of 2.84m. The spring has a spring constant of 250N/m. The crate is released and slides across a rough surface with a kinetic friction coefficient of 0.38 for a distance of 62m before colliding with and sticking to another crate. The second crate is filled with cans of smoked turkey and has a mass of 2kg. The two crates then slide horizontally off a vertical cliff with a height of 15m and land on the level ground below. Use g -9.8m/s^2, and answer the following:(no picture needed)

a) what maximum acceleration does the crate of stuffed kittens achieve due to the spring?

b) How fast is the crate of stuffed kittens moving when it collides with the crate of smoked turkey?

c) how fast are the two crates moving when they launch horizontally off the cliff and how fast are the two crates moving just before they hit the ground?

d) How much farther would the spring need to be compressed in order to double the final speed of the two crates at the moment they are about to hit the ground?

Solutions

Expert Solution

a) spring force=k. x

m. a = k.x

a= kx/m= 250×2.84/1 = 710 m/s-2

b) the enrgy stored due to spring compression =1/2kx^2 =250×2.84×2.84 /2 = 1008.2 J

Enrgy lost in doing work against friction

= Distance travelled× friction coefficient ×mass ×g

= 62×0.38×1×9.8 = 230.88 J

Thus the kinetic enrgy of kitten crate before hitting thecrate of smoked turkey = 1/2 mv^2 = 1008.2 - 230.9

1/2mv^2 = 777.3 J

v= 39.4 m/s

c) after the collision the two crates moves together. Applying law of momentum conservation we get=

mv= (m+M) V. Where M = 2kg

V = 1×39.4/ 3 = 13.13m/s

V is the velocity with which both the crate moves after collision.

Now the kinetic enrgy of both crate + potential enrgy = total kinetic energy

1/2(m+M) V^2 + (m+M) g h = 1/2 (m+M) v^2

3×13.13×13.13 /2 + 3× 9.8 ×15 = 3×v^2 /2

258.59 +441= 1.5 v^2

v = 21.6 m/s

d) two get the final speed of crate doubled that is 43.2 m/s the spring should be compressed by the longer distance that will be calculated using the above method reverse.

After calculation.... x = 7.39 m


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