Question

A crate is given a push across a horizontal surface. The crate has a mass m, the push gives it an initial speed of 1.90 m/s, and the coefficient of kinetic friction between the crate and the surface is 0.120.

(a) Use energy considerations to find the distance (in m) the crate moves before it stops.

(b) Determine the stopping distance (in m) for the crate if its initial speed is doubled to 3.80 m/s.

Answer 1

a)
`s = 1.534\,m`, b)
`s = 6.135\,m`

Step-by-step explanation:

a) The energy equation for the crate is modelled after the Principle of Energy Conservation and Work-Energy Theorem. Changes in gravitational potential energy can be neglected due to the information of a horizontal surface:

`K_(A) = W_(loss)`

`(1)/(2)\cdot m \cdot v^(2) = \mu_(k) \cdot m \cdot g\cdot s`

The distance that crate needs to cover before stopping is:

`s = (v^(2))/(2\cdot \mu_(k)\cdot g)`

`s = ((1.90\,(m)/(s) )^(2))/(2\cdot (0.120)\cdot (9.807\,(m)/(s^(2)) ))`

`s = 1.534\,m`

b) The stopping distance is:

`s = ((3.80\,(m)/(s) )^(2))/(2\cdot (0.120)\cdot (9.807\,(m)/(s^(2)) ))`

`s = 6.135\,m`