Tim lifting a 34kg weight to a maximum height of 2.3m and then lowering it to the floor repeatedly does not result in any net work done. This means that Tim's muscles are not gaining or losing energy from the lifting and lowering motion.
Tim's repeated lifting and lowering of a 34kg weight to a maximum height of 2.3m can be analyzed in terms of work and energy.
1. Work done in lifting the weight: The work done is equal to the force applied multiplied by the distance over which the force is exerted. In this case, the force Tim exerts is equal to the weight of the object, which is the mass multiplied by the acceleration due to gravity (9.8 m/s^2). The distance over which the force is exerted is the height to which the weight is lifted. Therefore, the work done in lifting the weight is given by:
Work = Force x Distance = (34kg x 9.8 m/s^2) x 2.3m = 761.24 Joules.
2. Work done in lowering the weight: When Tim lowers the weight back to the floor, gravity is doing the work. The work done by gravity is equal to the negative of the work done in lifting the weight. Therefore, the work done in lowering the weight is -761.24 Joules.
3. Total work done in one repetition: The total work done in one repetition is the sum of the work done in lifting and lowering the weight, which is 761.24 Joules + (-761.24 Joules) = 0 Joules.
4. Total work done in 15 repetitions: Since the work done in each repetition is 0 Joules, the total work done in 15 repetitions is also 0 Joules.