Question

A force of 100 N acts uphill on a 4 kg mass which sits on a surface with μ = 0.7 inclined at 23 degrees. Find the acceleration.

Answer 1

as per FBD we can see that there are two forces opposite to our applied force F

1. Component of weight along the inclined plane
`mgsin\theta`

2. Friction force along the inclined plane
`F_f`

now in order to find the acceleration we will use Newton's II law

`F_(net) = ma`

here we can say that net force is given by

`F_(net) = F - mgsin\theta - F_f`

now we can find friction force by its formula

`F_f = \muN`

here normal force is

`N = mgcos\theta`

now we have

`F_f = \mu mgcos\theta`

now net force is given by

`F_(net) = F - mgsin\theta - \mu mg cos\theta`

now acceleration is given by

`a = (F - mgsin\theta - \mu mgcos\theta)/(m)`

now we will plug in all values in above expression

`a = (100 - 4*9.8sin23 - 0.7 *4*9.8*cos23)/(4)`

[tex]a = 14.85 m/s^2[