Question

A 5.00-kg box slides 4.00 m across the floor before coming to rest. What is the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

Answer 1

Step-by-step explanation:

According to the given situation,

work done = change in kinetic energy

As the formula of kinetic energy is as follows.

K.E =
`(1)/(2)mv^(2)`

Putting the given values into the above formula as follows.

K.E =
`(1)/(2)mv^(2)`

=
`(1)/(2) * 5 kg * (3 m/s)^(2)`

= 22.5 J

Also we know that,

Work done = force x distance

or, force =
`(work)/(distance)`

=
`(22.5 J)/(4 m)`

= 5.62 N

In the given case, force is friction and formula to calculate friction is as follows.

friction =
`\mu * m * g`

where,
`\mu` = the coefficient of friction

Putting the given values into the above formula as follows.

Friction =
`\mu * m * g`

5.62N =
`\mu * 5 * 9.8`

`\mu` = 0.114

Thus, we can conclude that the coefficient of kinetic friction between the floor and the given box is 0.114.