Question

Calculate the force (in N) the woman in the figure below exerts to do a push-up at constant speed, taking all data to be known to three digits. (You may need to use torque methods from a later chapter.) m = 56 kg0.88 m1.47 m N (b) How much work (in J) does she do if her center of mass rises 0.180 m? J (c) What is her useful power output (in W) if she does 21 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.)

Answer 1

A. Force = 400.2 N

B. Work done = 120 J

C. W = 48 W

Why is this correct?

a) Under equilibrium conditions:

`\[ F * 1.45 \, \text{m} = 68 \, \text{kg} * 9.81 \, \text{m/s}^2 * 0.87 \, \text{m} \]`

This yields
`\( F = 400.2 \, \text{N} \)`.

b) Work done:

`\[ \text{Work done} = m * g * h = 68 \, \text{kg} * 9.81 \, \text{m/s}^2 * 0.180 \, \text{m} = 120.0744 \, \text{J} = 120 \, \text{J} \]`

c) Power:

`\[ \text{Power} = 120.0744 \, \text{J} * \left((24)/(60) \, \text{s}\right) = 48.02976 \, \text{W} = 48 \, \text{W} \]`

Force, a fundamental concept in physics, signifies any interaction capable of altering the motion of an object if unopposed. It stands as a vector quantity, encompassing both magnitude and direction, and is measured in newtons (N).

Work, within the realm of physics, materializes when an applied force induces displacement in an object. Quantitatively, work (W) is the outcome of the product between the applied force (F) and the displacement (d) of the object, specifically aligned with the force's direction.

Power, on the other hand, embodies the pace at which work is accomplished or energy is transformed. It characterizes the ratio of work (W) to the time interval (t) over which the work is executed. This measure of power, expressed in watts (W), delineates the efficiency or swiftness of energy conversion or work completion.

See missing part of the question on the attached image.