Question

An applied force of 75.0 N accelerates a 10.0 kg block at 3.5 m/s2 along a horizontal surface. a. How large is the frictional force? b. What is the coefficient of friction?

Answer 1

a. 40.0 N

b. 0.41

Step-by-step explanation:

By the second law of Newton, the net force is equal to the mass times the acceleration. In this case, the net force is the applied force less the frictional force, so

`\begin{gathered} F_{\text{net}}=ma \\ F-F_f=ma \end{gathered}`

Where F is the applied force, Ff is the friction force, m is the mass and a is the acceleration.

Solving for Ff, we get

`Ff=F-ma`

Now, we can replace the values

So, the frictional force is 40N

On the other hand, the force of friction is equal to the normal force times the coefficient of friction. Where the normal is equal to its weight, so

`\begin{gathered} F_f=\mu F_n \\ F_f=\mu mg \end{gathered}`

Where μ is the coefficient of friction and g is the gravity and it is equal to 9.8 m/s². Solving for μ

`\mu=(F_f)/(mg)`

Now, we can replace the values to get

Therefore, the coefficient of friction is 0.41

Then, the answers are

a. 40.0 N

b. 0.41