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swat32
3 years ago
12

Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco

sθ). Note that the acceleration is independent of mass and reduces to the expression found in the previous problem when friction becomes negligibly small (μk=0).

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

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An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
andrew-mc [135]

Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

  • x0 is the initial position.
  • v0 is the initial velocity.
  • a is the acceleration.
  • t is the time interval in which the motion is studied.

In this case:

  • x0= 0
  • v0= 0  because the object is initially stationary
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  • t= 15 s

Replacing:

x= 0 + 0*15 s + ½*6 \frac{m}{s^{2} }*(15s)²

Solving:

x=½*6 \frac{m}{s^{2} }*(15s)²

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