In Fig. 6-54, the coefficient of kinetic friction between the block and inclined plane is 0.20, and angle θis 60°. What are the

(a) magnitude aand

(b) direction (up or down the plane) of the block’s acceleration if the block is sliding down the plane?

What are (c) aand (d) the direction if the block is sent sliding up the plane?

Short Answer

Expert verified
  1. a=7.5m/s2.
  2. The direction of the accelerationa is down the slope.
  3. a=9.5m/s2.
  4. The direction is down the slope.

Step by step solution

01

Given data

  • The angle of inclination of the ramp, θ=60°.
  • The coefficient of kinetic friction, μk=0.2.
02

To understand the concept

The problem deals with Newton’s second law of motion, which states that the acceleration a, of an object is dependent upon the net force F acting upon the object and the mass m, of the object when the system is in equilibrium.Use Newton’s law of motion to calculate the kinetic friction.

Formula:

F=ma

03

(a) Calculate the magnitude a (up or down the plane) of the block’s acceleration if the block is sliding down the plane.

The normal force on the block on the inclined ramp is:

FN=mgcosθ

To calculate acceleration, use Newton’s second law of motion:

ma=mgsinθ-fkma=mgsinθ-μkmgcosθ

Substitute the values in the above expression, and we get,

a=9.8m/s2×sin60°-0.2×9.8m/s2×cos60°a=7.5m/s2

Thus, the magnitude of a is 7.5m/s2.

04

(b) Calculate the direction (up or down the plane) of the block’s acceleration if the block is sliding down the plane.

From the above calculation, we can conclude that the direction of the accelerationa is down the slope.

05

(c) Calculate the magnitude a (up or down the plane) of the block’s acceleration if the block is sliding up the plane.

When the friction force is the downhill direction, then the equation of motion will be,

a=gsinθ+μkgcosθ

Substitute the values in the above expression, and we get,

a=9.8m/s2×sin60°+0.2×9.8m/s2×cos60°a=9.5m/s2

Thus, the magnitude of a is 9.5m/s2.

06

(d) Calculate the direction of a (up or down the plane) of the block’s acceleration if the block is sliding up the plane.

From the above calculation, we can conclude that the direction is down the slope.

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

Repeat Question 1 for force F angled upward instead of downward as drawn.

An airplane is flying in a horizontal circle at a speed of 480km/h(Fig. 6-41). If its wings are tilted at angleθ=40°to the horizontal, what is the radius of the circle in which the plane is flying? Assume that the required force is provided entirely by an “aerodynamic lift” that is perpendicular to the wing surface.

In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. (a) Suppose the slope angle isθ=40.0.The snow is dry snow with a coefficient of kinetic frictionμk=0.0400, the mass of the skier and equipment ism=85.0kg, the cross-sectional area of the(tucked) skier isA=1.30m2, the drag coefficient isc=0.150, and the air density islocalid="1654148127880" 1.20kg/m3. (a) What is the terminal speed? (b) If a skier can vary C by a slight amountdCby adjusting, say, the hand positions, what is the corresponding variation in the terminal speed?

In Fig. 6-24, a forceacts on a block weighing 45N.The block is initially at rest on a plane inclined at angle θ=150to the horizontal. The positive direction of the x axis is up the plane. Between block and plane, the coefficient of static friction is μs=0.50and the coefficient of kinetic friction is μk=0.34. In unit-vector notation, what is the frictional force on the block from the plane when is (a) (-5.0N)i^, (b) (-8.0N)i^, and (c) (-15N)?

In 1987, as a Halloween stunt, two skydivers passed a pumpkin back and forth between them while they were in free fall just west of Chicago. The stunt was great fun until the last skydiver with the pumpkin opened his parachute. The pumpkin broke free from his grip, plummeted about0.5 km, and ripped through the roof of a house, slammed into the kitchen floor, and splattered all over the newly remodeled kitchen. From the sky diver’s viewpoint and from the pumpkin’s viewpoint, why did the skydiver lose control of the pumpkin?

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