Question

A 24.5 kg child pulls a 3.75 kg toboggan up a hill inclined at 25.7° to the horizontal. The vertical height of the hill is 30.3 m. Friction is negligible. (a) Determine how much work the child must do on the toboggan to pull it at a constant velocity up the hill. (b) Now suppose that the hill is inclined at an angle of 19.6° but the vertical height is still 30.3 m. What conclusion can you make?

Answer 1

Given:

The mass of the child is,

`M=24.5\text{ kg}`

The mass of the toboggan is,

`m=3.75\text{ kg}`

The angle of the hill with the horizontal is,

`h=30.3\text{ m}`

To find:

a) The work the child must do on the toboggan to pull it at a constant velocity up the hill

b) Now suppose that the hill is inclined at an angle of 19.6° but the vertical height is still 30.3 m. What conclusion can you make?

Step-by-step explanation:

The free-body diagram is shown below:

As the object has constant velocity, the object is in equilibrium.

So,

`\begin{gathered} F_x=mgsin\theta \\ and \\ R=mgcos\theta \end{gathered}`

The friction is negligible.

So,

`f=\mu R=0`

The work done by the child is,

`\begin{gathered} W=F_x* displacement\text{ along the slope} \\ =mgsin\theta*(h)/(cos\theta) \\ =mghtan\theta \\ =3.75*9.8*30.3tan\text{ 25.7}\degree \\ =535.9\text{ J} \end{gathered}`

Hence, the work is 535.9 J.

b)

Now for the given angle, the work is,

`\begin{gathered} W^(\prime)=3.75*9.8*30.3tan19.6\degree \\ =396.5\text{ J} \end{gathered}`

The work depends on the angle of inclination, the work is less for the lower angle of inclination of the hill.