Question

When a 40-N force, parallel to the incline and directed up the incline, is applied to a crate on a frictionless incline that is 30° above the horizontal, the acceleration of the crate is 2.0 m/s^2, up the incline. The mass of the crate is ?

Answer 1

`m ~= 5.8 kg`

Explanation:

The net force up the incline is given by

`F_"net" = m*a`

`F_"net"` is the sum of the 40 N force up the incline and the component of the object's weight, `m*g`, down the incline.

`F_"net" = 40 N - m*g*sin30 = m*2 m/s^2`

Solving for m,

`m*2 m/s^2 + m*9.8 m/s^2*sin30 = 40 N`

`m*(2 m/s^2 + 9.8 m/s^2*sin30) = 40 N`

`m*(6.9 m/s^2) = 40 N`

`m = (40 N)/(6.9 m/s^2)`

Note: the Newton is equivalent to `kg*m/s^2`. (Refer to F=ma to confirm this.)

`m = (40 kg*cancel(m/s^2))/(4.49 cancel(m/s^2)) = 5.8 kg`

I hope this helps,
Steve

Answer 2

`5.793\ kg`

Explanation:

Given that a force `F=40 \N` is applied on the crate of mass `m` kg to cause it to move with an acceleration `a=2\ \text{m/s}^2` up the plane inclined at an angle `\theta=30^\circ` with the horizontal.

Applying Newton's second law , the net force acting on the crate moving up the inclined plane

`F_{\text{net}}=ma`

`F-mg\sin\theta=ma`

`F=m(a+g\sin\theta)`

`m=\frac{F}{a+g\sin\theta}`

`=\frac{40}{2+9.81\sin30^\circ}`

`=\frac{40}{6.905}`

`=5.793\ kg`