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Mashutka [201]
3 years ago
13

A 2.8 kg block slides with a speed of 2.4 m/s on a frictionless horizontal surface until it encounters a spring. Part A If the b

lock compresses the spring 5.6 cm before coming to rest, what is the force constant of the spring
Physics
1 answer:
Fudgin [204]3 years ago
5 0

Answer:

<h2>5,142.86</h2>

Explanation:

The kinetic energy possessed by the block when sliding will be equal to the energy needed to compress the string.

Kinetic energy = 1/2 mv² and energy stored in the spring = 1/2 ke²

m = mass of the block (in kg) = 2.8 kg

v = speed of the block (in m/s) = 2.4 m/s

k = force constant of the spring

e = extension (in metres) = 0.056m

Since KE = energy stored in the spring

1/2 mv² = 1/2 ke²

mv² = ke²

2.8(2.4)²  = k(0.056)²

16.128 = 0.003136k

k = 16.128/0.003136

k =  5,142.86

The force constant of the spring is 5,236.36

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