A 40 N block lifted straight upward by a hand applying a force of 40 N has an initial kinetic energy of 12J. If the block is lifted 0.5 m, how mauch work does the hand do? What is the block's final kinetic energy?
work=force*distance
FinalKe=InitialKE+workdone-changePE
Looks like the finalKE is the same as initial KE.
To solve this problem, we need to understand the concepts of work, kinetic energy, and force. Work is defined as the product of force and displacement, and kinetic energy is the energy possessed by an object due to its motion.
1. First, let's calculate the work done by the hand:
Work = Force * Displacement
In this case, the force applied by the hand is 40 N, and the displacement is 0.5 m:
Work = 40 N * 0.5 m
= 20 J
Therefore, the hand does 20 J of work on the block.
2. Next, let's find the final kinetic energy of the block.
We know that the initial kinetic energy is 12 J.
The work done on an object is equal to the change in its kinetic energy:
Work = Change in Kinetic Energy
Since the kinetic energy initially was 12 J and the work done by the hand is 20 J, the change in kinetic energy can be calculated as follows:
Change in Kinetic Energy = Work
= 20 J
Therefore, the final kinetic energy of the block is the sum of the initial kinetic energy and the change in kinetic energy:
Final Kinetic Energy = Initial Kinetic Energy + Change in Kinetic Energy
= 12 J + 20 J
= 32 J
Thus, the block's final kinetic energy is 32 J.