Let's go through each part of the problem step by step.

Part (a): Calculate the Force

To calculate the force the woman exerts during a push-up, we need to consider her body as a rigid object and apply the principles of torque (moment of force). Here's how you can approach it:

Setup:

• $m = 53 \, \text{kg}$: Mass of the woman.
• $g = 9.81 \, \text{m/s}^2$: Acceleration due to gravity.
• $d_1 = 0.86 \, \text{m}$: Distance from her feet to her center of mass.
• $d_2 = 1.43 \, \text{m}$: Total length from her feet to her hands.

Let's assume that her hands and feet exert forces $F_h$ and $F_f$ respectively, and that the body is in equilibrium (no rotation).

Torques about the Feet:

Taking torques about the feet, we have:

$F_h \cdot d_2 = m \cdot g \cdot d_1$

$F_h = \frac{m \cdot g \cdot d_1}{d_2}$

Now substitute the values:

$F_h = \frac{53 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.86 \, \text{m}}{1.43 \, \text{m}}$

Calculating the force:

$F_h = \frac{446.97 \, \text{N} \times 0.86 \, \text{m}}{1.43 \, \text{m}} = \frac{384.4 \, \text{N} \cdot \text{m}}{1.43 \, \text{m}} \approx 268.9 \, \text{N}$

So, the force exerted by her hands (in $\text{N}$) is approximately 268.9 N.

Part (b): Calculate the Work Done

Work done by the woman when her center of mass rises by 0.260 m can be calculated using the work-energy principle:

$W = F \times d$

Where:

• $F = m \cdot g$: The force due to gravity (weight of the woman).
• $d = 0.260 \, \text{m}$: The vertical displacement of her center of mass.

Now substitute the values:

$W = 53 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 0.260 \, \text{m}$

$W = 135.08 \, \text{N} \times 0.260 \, \text{m} = 135.08 \, \text{J}$

So, the work done is approximately 135.1 J.

Part (c): Calculate the Useful Power Output

Power is the rate at which work is done. For 30 push-ups in 1 minute:

• Number of push-ups = 30
• Time = 1 minute = 60 seconds
• Work done per push-up = $W$

Total work done in 30 push-ups:

$W_{\text{total}} = 30 \times 135.1 \, \text{J} = 4053 \, \text{J}$

Now, power $P$ is given by:

$P = \frac{W_{\text{total}}}{t}$

Substitute the values:

$P = \frac{4053 \, \text{J}}{60 \, \text{s}} \approx 67.55 \, \text{W}$

So, her useful power output is approximately 67.55 W.

Summary of Answers:

(a) Force exerted: 268.9 N

(b) Work done: 135.1 J

(c) Power output: 67.55 W

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Would you like to go into more details on any part?

Questions:

1. How is torque used in calculating the force in part (a)?
2. Why is gravity involved in the calculation of work done?
3. What is the significance of calculating power output in part (c)?
4. How would the force change if the distance from her feet to her hands was increased?
5. How does the number of push-ups affect the total work done?
6. What assumptions are made about the woman's body in this calculation?
7. How would the work done change if the rise in her center of mass was 0.5 m instead of 0.26 m?
8. Why is it unnecessary to include work done during lowering in the power calculation?

Tip: Always double-check the units when solving physics problems to ensure your answers are in the correct form.