Question

A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface.

a) What would be the block's acceleration be if the surface were frictionless?


b) How large is the kinetic friction force?


c) What is the coefficient of kinetic friction?

Answer 1

(a). The block's acceleration be if the surface were friction less is 8 m/s².

(b). The kinetic friction force is 10 N.

(c). The coefficient of kinetic friction is 0.2

Step-by-step explanation:

Given that,

Force = 40.0 N

Acceleration = 6.0 m/s²

Mass = 5.0 kg

(a). We need to calculate the block's acceleration be if the surface were friction less

Using formula of force

`F= ma`

Where, m = 5.0 kg

F = 40.0 N

`40.0= 5.0* a`

`a=(40.0)/(5.0)`

`a=8\ m/s^2`

(b). We need to calculate the kinetic friction force

Using formula of frictional force

`f_(k)=F-ma`

Put the value into the formula

`f_(k)=40-5.0*6.0`

`f_(k)=10\ N`

(c). We need to calculate the coefficient of kinetic friction

Using formula of friction formula

`f_(k)=\mu mg`

`\mu=(f_(k))/(mg)`

`\mu=(10)/(5.0*9.8)`

`\mu=0.2`

Hence, (a). The block's acceleration be if the surface were friction less is 8 m/s².

(b). The kinetic friction force is 10 N.

(c). The coefficient of kinetic friction is 0.2

Answer 2

A force of 40.0 N accelerating a 5.0-kg block at 6.0
`m/s^2` along a horizontal surface has the following parameters:

a. Acceleration = 8
`m/s^2`

b. Kinetic friction force, Fk = 10 Newton.

c. Coefficient of kinetic friction, u = 0.2040.

Given the following data:

Force = 40.0 Newton

Acceleration = 6.0
`m/s^2`

Mass of block = 5 kg

a. To find the block's acceleration if the surface were frictionless, we would apply Newton's Second Law of Motion:

`Acceleration = (Force)/(Mass)\\\\Acceleration = (40)/(5)`

Acceleration = 8
`m/s^2`

b. To find the kinetic friction force, we would apply the frictional force formula:

`F_k = Force - (ma)\\\\F_k = 40 - (5[6])\\\\F_k = 40 - 30`

Kinetic friction force, Fk = 10 Newton

c. To find the coefficient of kinetic friction:

`u = (F_k)/(mg)\\\\u = (10)/(5[9.8])\\\\u = (10)/(49)`

Coefficient of kinetic friction, u = 0.2040.