From a 5 metre high smooth hill a 6 kg block slides down and stops somewhere on the horizontal ground. Effect of turning part is negligible. Find the work done while the block is pulled back from this position to the top of the hill.
600 j
To find the work done while pulling the block back from its position on the ground to the top of the hill, we need to consider the changes in potential energy.
The potential energy of an object is given by the equation:
PE = m * g * h
Where:
m = mass of the object (6 kg in this case)
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height (in meters)
Given that the height of the hill is 5 meters, the potential energy of the block at the top of the hill would be:
PE_top = 6 kg * 9.8 m/s^2 * 5 m
Now, when the block is on the ground, its potential energy becomes zero since the reference point for potential energy is usually taken at the ground level. Therefore, the change in potential energy is calculated as:
ΔPE = PE_top - PE_ground
ΔPE = 6 kg * 9.8 m/s^2 * 5 m - 0
So, the work done while pulling the block back to the top of the hill is equal to the change in potential energy, which is:
Work = ΔPE
Work = 6 kg * 9.8 m/s^2 * 5 m
By performing the calculations, the work done in pulling the block back to the top of the hill is found to be:
Work = 294 J (Joules)