Question

an ideal spring is usedto fire a 15g block horizontally across a frictionless table top. the spring has a spring constant of 20N/m and it initially compressed by 7.0cm. the speed of the block as it leaves the spring is

Answer 1

Potential energy of the spring = kinetic energy of the block

1/2kx^2 = 1/2mv^2

v = √k/mx=√20 × 1000/15×0.07=2.55 m/s

Answer 2

Concept:

The potential energy of a spring is given by,

P = 1/2 kx²

Where P = Potential energy

k = Spring Constant

x = Distance moved by spring

The kinetic energy of an object is given by,

K = 1/2 mv²

Where K = Kinetic energy

m = Mass of the object

v = Velocity of the object

Given:

Mass of the block, m = 15 g = 0.015 kg

Spring constant, k = 20 N/m

Distance moved by spring, x = Distance to which it is compressed = 7 cm = 0.07 m

Find:

The speed of the block as it leaves the spring.

Answer:

The speed of the block as it leaves the spring is 2.556 m/s.

Solution:

The potential energy of the spring is given by,

P = 1/2 kx²

Putting the values in the above equation, we get

P = 1/2 (20)(0.07)²

P = 10(0.0049)

P = 0.049 J

Let the speed of the block be v.

The kinetic energy of the block is given by,

K = 1/2 mv²

K = 1/2 (0.015) v²

K = 0.0075 v²

Now, as the block moves due to the spring (table is frictionless), according to the law of conservation of energy, we have

P = K

0.049 = 0.0075 v²

v² = 0.049/0.0075

v² = 6.53

v = √6.53

v = 2.556 m/s

Hence, the speed of the block as it leaves the spring is v = 2.556 m/s.

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