Question

A trunk w/ a mass of 100kg is pulled 10m across a floor by a rope making an angle of 30 degree w/ the horizantal.if the coeffient of sliding frictions is 0.2.find
a. the force of the rope
b.the work done

Answer 1

First of all, make a force diagram of the trunk.
The friction force is \[f_k = \mu \times n\] where n is the normal to the weight (Newton's 3rd Law, equal and opposite force).

Now, the trunk is not accelerating, this means the sum of the forces acting on the trunk must be zero.

Write the equation for the sum of horizontal forces:
\[f_k = Fcos(30)\]
From this, you can work out F in terms of n.

To work out the normal, write the equation for the sum of the vertical forces acting on the trunk:
\[n+Fsin(30)=mg\]

Use the value for the normal in your first equation, and then you have your force from the rope.

Answer 2

The work done will be the scalar product of the force acting on the object (F) and the distance moved (s).
Work out the horizontal component of the force of the string:
\[w=Fscos(30)\]