Question

A 50Kg desk is pushed across the floor, at a constant rate, with a force of 160N. What is the magnitude of the kinetic friction coefficient between the desk and the floor?

Answer 1

Step-by-step explanation:

If the desk is not accelerating , but only moving at a constant velocity . the push force is JUST equal to the kinetic friction force = 160 N

Answer 2

The magnitude of the kinetic friction coefficient between the desk and the floor is approximately
`\( 0.3265 \)`.

To find the kinetic friction coefficient
`(\( \mu_k \))`, you can use the formula:

`\[ \text{Force of Kinetic Friction} = \mu_k * \text{Normal Force} \]`

The normal force
`(\( F_N \))` is the force exerted by a surface that is perpendicular to the surface. When an object is on a horizontal surface and is not accelerating vertically (i.e., it's not lifting off or sinking into the surface), the normal force is equal to the weight of the object. The weight
`(\( W \))` is given by:

`\[ W = m * g \]`

where
`\( m \)` is the mass of the object and
`\( g \)` is the acceleration due to gravity (approximately
`\( 9.8 \ m/s^2 \)`.

In this case:

`\[ m = 50 \ kg \]`

`\[ g = 9.8 \ m/s^2 \]`

`\[ W = 50 \ kg * 9.8 \ m/s^2 = 490 \ N \]`

Now, we can use the force of kinetic friction
`(\( F_{\text{friction}} \))` and the normal force to find the kinetic friction coefficient:

`\[ F_{\text{friction}} = \mu_k * F_N \]`

Given that
`\( F_{\text{friction}} = 160 \ N \)` and
`\( F_N = 490 \ N \)`, you can rearrange the formula to solve for
`\( \mu_k \)`:

`\[ \mu_k = \frac{F_{\text{friction}}}{F_N} \]`

`\[ \mu_k = (160 \ N)/(490 \ N) \]`

`\[ \mu_k \approx 0.3265 \]`

So, the magnitude of the kinetic friction coefficient between the desk and the floor is approximately
`\( 0.3265 \)`.