Question

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is 2.4 N. The free-body diagram shows the forces acting on the sled.

What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?

1) a = 1.3 m/s2; FN = 63.1 N
2) a = 1.6 m/s2; FN = 65.6 N
3) a = 1.9 m/s2; FN = 93.7 N
4) a = 2.2 m/s2; FN = 78.4 N

Answer 1

1) a = 1.3 m/s2; FN = 63.1 N

Step-by-step explanation:

edge 2021

Answer 2

As per the question the mass of sled[m] is given as 8 kg.

The fictional force
`[F_(f) ]` is given as 2.4 N

A force
`[F_(p) ]` of 20 N was exerted on the sled at angle of
`50^(0)`

Resolving the force into horizontal and vertical components we get -

`F_(h) =F_(p) cos\theta` and
`F_(v) =F_(p) sin\theta`

Here
`F_(h)` is the horizontal component and
`F_(v)` is the vertical component.

`F_(h) =20*cos50`
`=12.85575219 N`

Similarly
`F_(v) =20*sin50=15.32088886 N`

From the free body diagram we get that sum of vertical components is zero as there is no motion in vertical direction.

Hence
`F_(N) +F_(p) sin\theta=mg` where g is the acceleration due to gravity .

⇒
`F_(N) =mg-F_(p) sin50`

⇒
`F_(N) =8*9.8-15.32088886` [here value of g is 9.8 m/s^2]

=63.1 N

The net motion of the body is along the forward direction.

Hence
`F_(p) cos\theta -F_(f) =ma` where a is the acceleration of the body.

⇒
`a=(F_(p)cos\theta -F_(f)  )/(m)`

`=(12.85575219 -2.4)/(8)`

`=1.30 m/s^2`

Hence the option A is the right answer.