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Chapter 4: Problem 22

A 2010 -kg elevator moves with an upward acceleration of $1.50 \mathrm{m} / \mathrm{s}^{2} .$ What is the force exerted by the cable on the elevator?

Short Answer

Expert verified
Answer: The force exerted by the cable on the elevator is 22716.1 N upward.

Step by step solution

01

Identify known values

We are given the mass of the elevator (m = 2010 kg) and its upward acceleration (a = 1.50 m/s²). The gravitational acceleration (g) is approximately 9.81 m/s² downward.
02

Calculate gravitational force

The gravitational force (Fg) is the force of gravity acting on an object, which is equal to the mass of the object multiplied by the gravitational acceleration. In this case, the gravitational force on the elevator is: Fg = m * g Fg = 2010 kg * 9.81 m/s² Fg = 19701.1 N (downward)
03

Apply Newton's Second Law

According to Newton's Second law, the net force (F_net) acting on an object is equal to its mass multiplied by its acceleration: F_net = m * a In this case, the net force acting on the elevator is the difference between the force exerted by the cable (Fc) and the gravitational force (Fg): F_net = Fc - Fg (since Fg is downward, and Fc is upward) We also know that F_net = m * a, so we can substitute that in the equation: m * a = Fc - Fg
04

Solve for force exerted by the cable

Now, we can solve for the force exerted by the cable (Fc) by plugging in the known values and solving for Fc: 2010 kg * 1.50 m/s² = Fc - 19701.1 N 3015 N = Fc - 19701.1 N Add 19701.1 to both sides of the equation to isolate Fc: Fc = 3015 N + 19701.1 N Fc = 22716.1 N
05

Final answer

The force exerted by the cable on the elevator is 22716.1 N (upward).

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Most popular questions from this chapter

A 10.0 -kg block is released from rest on a frictionless track inclined at an angle of \(55^{\circ} .\) (a) What is the net force on the block after it is released? (b) What is the acceleration of the block? (c) If the block is released from rest, how long will it take for the block to attain a speed of \(10.0 \mathrm{m} / \mathrm{s} ?\) (d) Draw a motion diagram for the block. (e) Draw a graph of \(v_{x}(t)\) for values of velocity between 0 and $10 \mathrm{m} / \mathrm{s}\(. Let the positive \)x$ -axis point down the track.
A parked automobile slips out of gear, rolls unattended down a slight incline, and then along a level road until it hits a stone wall. Draw an FBD to show the forces acting on the car while it is in contact with the wall.
A skydiver, who weighs \(650 \mathrm{N},\) is falling at a constant speed with his parachute open. Consider the apparatus that connects the parachute to the skydiver to be part of the parachute. The parachute pulls upward with a force of \(620 \mathrm{N} .\) (a) What is the force of the air resistance acting on the skydiver? (b) Identify the forces and the interaction partners of each force exerted on the skydiver. (c) Identify the forces and interaction partners of each force exerted on the parachute.
A man lifts a 2.0 -kg stone vertically with his hand at a constant upward acceleration of \(1.5 \mathrm{m} / \mathrm{s}^{2} .\) What is the magnitude of the total force of the man's hand on the stone?
A 6.0 -kg block, starting from rest, slides down a frictionless incline of length \(2.0 \mathrm{m} .\) When it arrives at the bottom of the incline, its speed is \(v_{\mathrm{f}} .\) At what distance from the top of the incline is the speed of the block \(0.50 v_{\mathrm{f}} ?\)
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