Final answer:
The resultant work on the 3-kg block sliding down the inclined plane is 705.6 J. This is calculated by finding the work done by gravity and the work done by friction. The work done by gravity is determined using the formula W = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the vertical height. The work done by friction is calculated using the formula W = μkFn, where μk is the coefficient of kinetic friction and Fn is the normal force.
Step-by-step explanation:
To find the resultant work on the 3-kg block sliding down the inclined plane, we need to calculate the work done by gravity and the work done by friction. The work done by gravity is given by the formula W = mgh, where m is the mass of the block (3 kg), g is the acceleration due to gravity (9.8 m/s²), and h is the vertical height (20 m). So, the work done by gravity is W = (3 kg)(9.8 m/s²)(20 m) = 588 J.
The work done by friction can be calculated using the formula W = μkFn, where μk is the coefficient of kinetic friction (0.2) and Fn is the normal force. The normal force is equal to the weight of the block, which is given by the formula Fn = mg, where m is the mass of the block (3 kg) and g is the acceleration due to gravity (9.8 m/s²). So, the normal force is Fn = (3 kg)(9.8 m/s²) = 29.4 N. Therefore, the work done by friction is W = (0.2)(29.4 N)(20 m) = 117.6 J.
The resultant work on the block is the sum of the work done by gravity and the work done by friction. So, the resultant work is W = 588 J + 117.6 J = 705.6 J.