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Simora [160]
3 years ago
12

A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient

of kinetic friction of 0.173. (indicate the direction with signs of your answer) (a) What is the force of kinetic friction in N acting on the block? (b) What is the block's acceleration in /s^2? (c) How Far will it slide (in m) before coming to rest? Plz answer as soon as possible
Physics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer:

Explanation:

a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force

R = mg where m is mass of the block

Force of friction F = μ x mg

= .173 x 12.2 x 9.8

= 20.68 N

b ) Only force of friction is acting on the body so

deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²

acceleration = - 1.7 m /s²

c )

v² = u² - 2 a s

v = 0 , u = 3.9 m /s

a = 1.7 m /s

0 = 3.9² - 2 x 1.7 x s

s = 4.47  m

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A string is wrapped around a pulley with a radius of 2.0 cm and no appreciable friction in its axle. The pulley is initially not
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Read 2 more answers
How can i find the acceleration?(rope and grinder have no weight) *** sorry for my english
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The main formula to be used here is

                       Force = (mass) x (acceleration).

We'll get to work in just a second.  But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them.  So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.

On the <u>right</u> side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration.  I don't know what to do with that, because . . .

At the <em>same time</em>, both of those will be pulled <u>up</u> by the 10 kg on the other side
of the upper pulley.

I think I can handle the 10 kg, and work out the acceleration that IT has.

Let's look at only the forces on the 10 kg:

-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.

-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope. 

So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.

The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.

The acceleration of 10 kg, with 49 newtons of force on it, is

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3 years ago
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omeli [17]

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