Question

g=9.8m/s^2A student pulls his younger sister in a sled (combined mass 35 kg) with a rope that is angled at = 13° with a force of 270 N. The coefficient of friction between the sled and the ground is 0.27.(a) What is the normal force acting on the sled?(N)(b) What is the acceleration of the sled?(m/s2)

Answer 1

Given,

The combined mass of the sledge and the younger sister, m=35 kg

The angle made by the rope, θ=13°

The force applied, F=270 N

The coefficient of friction, μ=0.27

The acceleration due to gravity, g=9.8 m/s²

(a) The normal force is given by,

`N=mg-F\sin \theta`

On substituting the known values,

`\begin{gathered} N=35*9.8-270\sin (13^(\circ)) \\ =282.3\text{ N} \end{gathered}`

Thus the normal force is given by 282.3 N

(b) The net force on the sledge is

`\begin{gathered} ma=F\cos \theta-f \\ =F\cos \theta-N\mu \end{gathered}`

Where f is the frictional force.

Thus the acceleration will be,

`a=(F\cos \theta-N\mu)/(m)`

On substituting the known values,

`\begin{gathered} a=(270N*\cos 13^(\circ)-282.3*0.27)/(35) \\ =5.3m/s^2 \end{gathered}`

Thus the acceleration of the sledge is 5.3 m/s²