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Chapter 4: Problem 128

A woman of mass \(51 \mathrm{kg}\) is standing in an elevator. (a) If the elevator floor pushes up on her feet with a force of $408 \mathrm{N},$ what is the acceleration of the elevator? (b) If the elevator is moving at \(1.5 \mathrm{m} / \mathrm{s}\) as it passes the fourth floor on its way down, what is its speed 4.0 s later?

Short Answer

Expert verified
The acceleration of the elevator is approximately -1.80 m/s虏. The negative sign indicates that the elevator is accelerating downward. (b) What is the elevator's speed 4.0 s after passing the fourth floor? The elevator's speed 4.0 s later is approximately -5.70 m/s. The negative sign indicates that the elevator is moving downward.

Step by step solution

01

Calculate the woman's weight

To calculate the woman's weight, we will use the equation W = mg, where W is the weight, m is the mass (51 kg), and g is the acceleration due to gravity (9.8 m/s虏). \(W = 51 \times 9.8\) \(W = 499.8 \mathrm{N}\) The woman's weight is 499.8 N.
02

Calculate the net force acting on the woman

The net force acting on the woman is the difference between the force exerted by the elevator floor on her feet (408 N) and her weight (499.8 N). \(F_{net} = 408 - 499.8\) \(F_{net} = -91.8 \mathrm{N}\) The net force acting on the woman is -91.8 N. Negative sign indicates that the force is acting downward.
03

Calculate the acceleration of the elevator

Using Newton's second law of motion (F = ma), we can find the acceleration (a) of the elevator. \(a = \frac{F_{net}}{m}\) \(a = \frac{-91.8}{51}\) \(a \approx -1.80 \mathrm{m/s^2}\) The acceleration of the elevator is approximately -1.80 m/s虏. The negative sign indicates that the elevator is accelerating downward.
04

Calculate the elevator's speed 4.0 s later

Now, we need to find the elevator's speed 4.0 s after passing the fourth floor. We will use the equation of motion: v = u + at. We are given the initial velocity (u) as 1.5 m/s (moving downward), the acceleration (a) as -1.80 m/s虏, and the time (t) as 4.0 s. \(v = 1.5 + (-1.80) \times 4.0\) \(v = 1.5 - 7.20\) \(v = -5.70 \mathrm{m/s}\) The elevator's speed 4.0 s later is approximately -5.70 m/s. The negative sign indicates that the elevator is moving downward.

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Most popular questions from this chapter

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on the surface. The sanding block weighs \(2.0 \mathrm{N}\) and Brenda pushes on it with a force of $3.0 \mathrm{N}\( at an angle of \)30.0^{\circ}$ with respect to the vertical, and angled toward the wall. Draw an FBD for the sanding block as it moves straight up the wall at a constant speed. What is the coefficient of kinetic friction between the wall and the block?
A 320 -kg satellite is in orbit around the Earth \(16000 \mathrm{km}\) above the Earth's surface. (a) What is the weight of the satellite when in orbit? (b) What was its weight when it was on the Earth's surface, before being launched? (c) While it orbits the Earth, what force does the satellite exert on the Earth?
Refer to Example 4.19. What is the apparent weight of the same passenger (weighing \(598 \mathrm{N}\) ) in the following situations? In each case, the magnitude of the elevator's acceleration is $0.50 \mathrm{m} / \mathrm{s}^{2} .$ (a) After having stopped at the 15 th floor, the passenger pushes the 8 th floor button; the elevator is beginning to move downward. (b) The elevator is moving downward and is slowing down as it nears the 8 th floor.
A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of \(560 \mathrm{N}\) at an angle of \(15^{\circ}\) with the centerline of the canal. Find the sum of the two forces exerted by the horses on the barge.
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