a. How much force does an 80kgastronaut exert on his chair while sitting at rest on the launch pad?

b. How much force does the astronaut exert on his chair while accelerating straight up at 10m/s2?

Short Answer

Expert verified

a. Force exerted by the astronaut while sitting on chair at rest on the launch pad is 784N.

b. Force exerted by the astronaut while accelerating straight up is 1584N.

Step by step solution

01

Part (a) Step 1 : Given information

An 80kg astronaut is sitting on his chair while at rest on the launch pad.

02

Part (a) Step 2 : Calculation

The acceleration due to gravity is taken as, g=9.8m/s2.

Weight is the force that gravity will exert on him to keep him on the earth, it is the same force which he exerts on the chair.

And it is given by :-

F=m×gF=80kg×9.8m/s2F=784N

03

Part (a) Step 3 : Final answer

Force exerted by the astronaut while sitting on chair at rest on the launch pad is 784N.

04

Part (b) Step 4: Given information

The astronaut on chair is accelerating at10m/s2.

05

Part (b) Step 5 : Explanation

Acceleration while going straight up, a=10m/s2.

Here the acceleration will be due to net force of forces like gravity and the reaction force of the chair . Let it be R . Acceleration at which moves straight up by localid="1649439467778" 10m/s2.

Net force acting on him will be R - mg .

localid="1650265993057" R-mg=maR=m(g+a)R=80kg×(9.8m/s2+10m/s2)R=80kg×19.8m/s2R=1584N

06

Part (b) Step 6 : Final answer

Reaction Force of chair on the astronaut is 1584 N . Hence force on the chair by astronaut will also be 1584 N.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Block Ain FIGUREEX7.4is heavier than block Band is sliding down the incline. All surfaces have friction. The rope is massless, and the massless pulley turns on frictionless bearings. The rope and the pulley are among the interacting objects, but you’ll have to decide if they’re part of the system.

A very smart 3-year-old child is given a wagon for her birthday. She refuses to use it. “After all,” she says, “Newton’s third law says that no matter how hard I pull, the wagon will exert an equal but opposite force on me. So I will never be able to get it to move forward.” What would you say to her in reply?

While driving to work last year, I was holding my coffee mug in my left hand while changing the CD with my right hand. Then the cell phone rang, so I placed the mug on the flat part of my dashboard. Then, believe it or not, a deer ran out of the woods and on to the road right in front of me. Fortunately, my reaction time was zero, and I was able to stop from a speed of 20m/s in a mere 50m, just barely avoiding the deer. Later tests revealed that the static and kinetic coefficients of friction of the coffee mug on the dash are 0.50 and 0.30, respectively; the coffee and mug had a mass of 0.50kg; and the mass of the deer was 120kg. Did my coffee mug slide?

A 4.0kgbox is on a frictionless 35°slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.0kgweight. The picture for this situation is similar to FIGURE P7.39.

a. What is the tension in the string if the 4.0kgbox is held in place, so that it cannot move?

b. If the box is then released, which way will it move on the slope?

c. What is the tension in the string once the box begins to move?

A rope of lengthL and mass m is suspended from the ceiling. Find an expression for the tension in the rope at position y, measured upward from the free end of the rope.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free