Question

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is applied. The 100 kg block sits on a horizontal frictionless surface, but there is friction between the two blocks.

Answer 1

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Step-by-step explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

`F-f=ma`

Put the value into the formula

`340-f=60*2.0`

`f=340-60*2.0`

`f=220\ N`

We need to calculate the coefficient of friction

Using formula of friction force

`f= \mu mg`

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

`\mu=(220)/(60*9.8)`

`\mu =0.37`

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

`F = ma`

`a=(F)/(m)`

`a=(220)/(100)`

`a=2.2\ m/s^2`

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²