Question

A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.44 radians with the horizontal. The frictional coefficients between the crate and plane are (\muμs = 0.61, \muμk = 0.23 ). What is the magnitude of the acceleration (in meters/second2) of this crate as it slides up the incline?

Answer 1

The acceleration is
`a = 6.2 m/s^2`

Step-by-step explanation:

From the question we are told that

The angle which the inclined plane make with horizontal is
`\theta = 0.44 \ rad`

The frictional coefficients are
`\mu_(\mu s) = 0.61` and
`\mu_(\mu k) = 0.23`

The force acting on the crate is mathematically represented as

`f = F_w + F_N`

Here f is the net force at which the crate is sliding down the plane which is mathematically represented as

`f = ma`

`F_w` is the force due to weight which is mathematically represented as

`F_w = mg sin (\theta)`

and
`F_N` the force due to friction which is mathematically represented as

`F_N = \mu_(\mu k ) * mg cos(\theta )`

So

`ma = mgsin(\theta ) + \mu_(\mu k) mg cos(\theta )`

`a = gsin(\theta ) + \mu_(\mu k ) * g cos(\theta)`

substituting values

`a = 9.8 sin(0.44 ) + 0.23 * 9.8* cos(0.44)`

`a = 6.2 m/s^2`