Questions: The figure is a graph of the normal force exerted by the floor on a woman making a vertical jump. At what speed does she leave the ground? Hint: The force of the floor is not the only force imparting an impulse to the woman. (See image below) Express your answer using three significant figures. The graph shows just the normal forces, the jump time is 0.3 s , The normal force's impulse is postive and the weight's impulse is negative.

The figure is a graph of the normal force exerted by the floor on a woman making a vertical jump. At what speed does she leave the ground? Hint: The force of the floor is not the only force imparting an impulse to the woman. (See image below) Express your answer using three significant figures. The graph shows just the normal forces, the jump time is 0.3 s , The normal force's impulse is postive and the weight's impulse is negative.
Transcript text: The figure is a graph of the normal force exerted by the floor on a woman making a vertical jump. At what speed does she leave the ground? Hint: The force of the floor is not the only force imparting an impulse to the woman. (See image below) Express your answer using three significant figures. The graph shows just the normal forces, the jump time is 0.3 s , The normal force's impulse is postive and the weight's impulse is negative.
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Solution

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Solution Steps

Step 1: Calculate the impulse from the normal force.

The impulse from the normal force is the area under the force-time graph. The graph can be divided into a rectangle and a triangle.

  • Rectangle: Area = (0.2 s)(500 N) = 100 Ns
  • Triangle: Area = (1/2)(0.1 s)(2000 N) = 100 Ns
  • Total Impulse: 100 Ns + 100 Ns = 200 Ns
Step 2: Calculate the impulse due to the weight.

The weight acts downwards throughout the jump duration (0.3s). Assuming the normal force graph starts at _t_ = 0, weight acts for 0.3s. Let's represent the woman's mass as _m_ and the weight as _mg_. Then the impulse is:

  • Weight Impulse: - _mg_(0.3 s)
Step 3: Apply the impulse-momentum theorem.

The impulse-momentum theorem states that the change in momentum is equal to the net impulse. The net impulse here is the sum of the normal force impulse and the weight impulse. The initial vertical velocity is 0 m/s. Let _v_ be the final take-off velocity.

  • Net Impulse: 200 Ns - (0.3s)_mg = *mv_ - 0 200 - 0.3 * _mg_ = _mv_

Given the normal force balanced the weight initially when stationary, we know 500N = mg So: 200 Ns -0.3s * (500 N) = _m_v 50 Ns = mv

Step 4: Determine the mass (m).

Weight is _mg_ which is given as 500 N. Taking g = 9.8 m/s² Mass (_m_) = 500N / 9.8 m/s² ≈ 51 kg

Step 5: Calculate the take-off velocity (v).
  • 50 = 51 v
  • v = 50 / 51 ≈ 0.98 m/s

Final Answer:

The woman leaves the ground at a speed of 0.980 m/s.

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