What mechanical work must you do to lift a uniform log that is 3.5 m long and has a mass of 110 kg from the horizontal to a vertical position? [Hint: Use the work-energy principle.]

Mechanical work done to lift the log vertically is 1886.5 J

Explanation:

Consider that the radius of the log is very small compared to the length of the log and its center of mass along horizontal direction is lying to the ground. So, its initial potential energy is zero.

Since the log is uniform, its center of mass coincide with its geometric center.

According to the problem, when the log is vertically lifted, the work done is positive as the direction of force and displacement are same.

Applying Work-Energy theorem,

U₁ + W = U₂

Here U₁ and U₂ are initial and final potential energy of the log and W is the work done on the log to lift it vertically.

Since, U₁ is zero, so the above equation becomes:

W = U₂

The final potential energy, U₂ = (mgl)/2

Here m is mass of the log, l is length of the log and g is acceleration due to gravity.

So,

W = (mgh)/2

Substitute 110 kg for m, 3.5 m for l and 9.8 m/s² for g in the above equation.

W = 1886.5 J

your answer is c! hope this : )