Question

A crate resting on a horizontal floor (\muμs = 0.75, \muμk = 0.24 ) has a horizontal force F = 93 Newtons applied to the right. This applied force is the maximum possible force for which the crate does not begin to slide. If you applied this same force after the crate is already sliding, what would be the resulting acceleration (in meters/second2) ?

Answer 1

The acceleration is
`a = 5 \ m/s^2`

Step-by-step explanation:

From the question we are told that

The coefficient of kinetic friction is
`\mu_k = 0.24`

The coefficient of static friction is
`\mu_s = 0.75`

The horizontal force is
`F_h = 93 \ N`

Generally the static frictional force is mathematically represented as

`F_F = \mu_s * (m * g )`

The static frictional force is the equivalent to the maximum possible force for which the crate does not begin to slide So

`F_h = F_F = \mu_s * (m * g )`

=>
`93 = \mu_s * (m * g )`

=>
`m = (93)/(\mu_s * g )`

substituting values

`m = (93)/(0.75 * 9.8 )`

`m = 12.65 \ kg`

When the crate is already sliding the frictional force is

`F_s = \mu_k *(m * g )`

substituting values

`F_s = 0.24 * 12.65 * 9.8`

`F_s = 29.82 \ N`

Now the net force when the horizontal force is applied during sliding is

`F_(net) = F_h - F_s`

substituting values

`F_(net) = 93 - 29.8`

`F_(net) = 63.2 \ N`

This net force is mathematically represented as

`F_(net ) = m * a`

Where a is the acceleration of the crate

So

`a = (F_(net))/(m )`

`a = ( 63.2)/(12.65 )`

`a = 5 \ m/s^2`