Question

A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The spring has a force constant of 1040 N/m. The coefficient of kinetic friction between the block and the floor is 0.26. The block is released from rest. The speed of the block after it has traveled 0.020 m is

Answer 1

The velocity of block = 0.188
`(m)/(s)`

Step-by-step explanation:

Mass m = 5.6 kg

k = 1040
`(N)/(m)`

`\mu` = 0.26

`x_(1) =` 0.035 m ,
`v_(1)` = 0

`x_(2) =` 0.02 m

From work energy theorem

`K_(1) + U_(1) + W_(other) = K_(2) + U_(2)` --------- (1)

Kinetic energy

`K = (1)/(2) k x^(2)` ------- (1)

Potential energy

`U = (1)/(2) k x^(2)` ------- (2)

Work done

W = F.s ------ (3)

From Newton's second law

`R_(N)` = mg

`R_(N)` = 5.6 × 9.81 = 54.9 N

Friction force = 0.4 × 54.9 = 21.9 N

Now the work done by the friction

`W_(f)` = - 21.9 × 0.015

`W_(f)` = - 0.329 J

Now kinetic energy

At point 1

`K_(1) = (1)/(2) m v^(2) _(1)`

`K_(1) = 0`

`U_(1) = (1)/(2) k x^(2)`

`U_(1) = (1)/(2) (1040) 0.035^(2)`

`U_(1) =` 0.637 J

At point 2

`K_(2) = (1)/(2) (5.6) v^(2) _(2)`

`K_(2) = 2.8 v_(2) ^(2)`

Potential energy

`U_(2) = (1)/(2) k x_2^(2)`

`U_(2) = (1)/(2) (1040) 0.02^(2)`

`U_(2) = 0.208` J

From equation (1) we get

0 + 0.637 - 0.329 = 2.8
`v_(2) ^(2)` + 0.208

2.8
`v_(2) ^(2)` = 0.1

`v_(2) =` 0.188
`(m)/(s)`

This is the velocity of block.