/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q38P A 40聽KGskier skis directly down... [FREE SOLUTION] | 91影视

91影视

A 40KGskier skis directly down a frictionless slope angled at 100to the horizontal. Assume the skier moves in the negative direction of an xaxis along the slope. A wind force with component role="math" localid="1657169142011" Fxacts on the skier. (a) What is Fxif the magnitude of the skier鈥檚 velocity is constant, (b) What isif the magnitude of the skier鈥檚 velocity is increasing at a rate of 1.0m/s2, and (c) What is Fxif the magnitude of the skier鈥檚 velocity is increasing at a rate of2.0m/s2?

Short Answer

Expert verified

(a) The x component of wind force when skier moves with constant velocity is, fx=68N(uphill).

(b) The x component of wind force when skier moves with the acceleration of role="math" localid="1657169523821" a=1.0m/s2is,Fx=28N(uphill).

(c) The x component of wind force when skier moves with the acceleration of a=2.0m/s2is,Fx=-12N (downhill).

Step by step solution

01

Given data

  • The mass of the skier is, m=40kg.
  • The inclination angle with the horizontal is, =100.
  • The acceleration of a skier is, a=1.0m/s2.
  • The acceleration of the skier is,a=2.0m/s2.
02

Understanding the concept of Newton’s second law

As per Newton鈥檚 second law, the force acting on the object is directly proportional to the acceleration of the body. The force is equal to the product of mass and acceleration of the body. A free-body diagram shows the forces that are acting on the body along with their directions.

By using Newton鈥檚 second law x component of wind force when skier moves with constant velocity and with acceleration can be found.

Formulae:

Fnet=ma

Here,Fnet is the net force, m is mass anda is acceleration.

03

Draw the free body diagram

04

a) Calculate Fx when skier moves with constant velocity

By applying Newton鈥檚 second law along x axis as,

Fnet=ma

The skier is moving with constant velocity hence it has zero acceleration. Consider the sign convention according to the direction of skier

Fx-mg.sin=0Fx=mg.sin=40kg9.8m/s2sin(100)=68N(uphill)

Hence, the value of role="math" localid="1657170901844" Fxwhen skier is moving with constant velocity is 68N .

05

b) Calculate Fx when skier moves with acceleration 1.0m/s2

The skier is going down with some acceleration, hence according to the Newton鈥檚 second law,

Fx-mg.sin=-maFx=mg.sin-ma=40kg9.8m/s2sin(100)-40kg1.0m/s2=28N(uphill)

Hence, the value of Fxwhen skier is going down with 1.0m/s2 the acceleration is28N.

06

c) Calculate Fx when skier moves with acceleration 2.0m/s2

The skier rate is increasing in downward direction, hence according to the Newton鈥檚 second law,

Fx-mg.sin=-maFx=mg.sin-maFx=(40kg9.8m/s2sin(100))-(40kg1.0m/s2)Fx=-12N(downhill)

Hence, the value of Fxwhen skier is going down with the 2.0m/s2acceleration is -12 N

The negative sign indicates that wind is also flowing in the downward direction, hence skier rate increases.

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 91影视!

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

A block is projected up a frictionless inclined plane with initial speedv0=3.50m/sThe angle of incline is=32.0. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?

A 鈥渟un yacht鈥 is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of 900 kgand receives a push of 20 N.

(a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

A firefighter who weighs 712 N slides down a vertical pole with an acceleration of 3.00 m/s2, directed downward. (a) What is the magnitude of the vertical force on the firefighter from the pole and (b) What is the direction (up or down) of the vertical force on the firefighter from the pole and (c) What is the magnitude of the vertical force on the pole from the firefighter? and (d) What is the direction of the vertical force on the pole from the firefighter?

You pull a short refrigerator with a constant force F鈬赌across a greased (frictionless) floor, either withF鈬赌horizontal (case 1) or with tilted upward at an angle(case 2). (a)What is the ratio of the refrigerator鈥檚 speed in case 2 to its speed in case 1 if you pull for a certain time t? (b) What is this ratio if you pull for a certain distance d?

Tarzan, who weighs820N , swings from a cliff at the end of a 20.0mvine that hangs from a high tree limb and initially makes an angle of 22.0with the vertical. Assume that an xaxis extends horizontally away from the cliff edge and a yaxis extends upward. Immediately after Tarzan steps off the cliff, the tension in the vine is760N . Just then, (a) what is the force on him from the vine in unit-vector notation and the net force on him (b) what is in unit-vector notation and as (c) what is a magnitude and (d) what is the angle relative to the positive direction of the xaxis? (e) What is the magnitude of Tarzan鈥檚 acceleration and (f) What is the angle of Tarzan鈥檚 acceleration just then?

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.