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andreev551 [17]

1 year ago

15

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object

is 4.2

1 answer:

Ann [662]1 year ago

3 0

The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object is 16.4 N.

<h3>What is frictional force?</h3>

Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.

The normal force acting on the object of mass 4.2 Kg is N = mg

N = 4.2 Kg × 9.8 m/s² = 41.16 N

Frictional force = ц N

= 0.40 × 41.16 N

= 16.4 N.

Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N

To find more on frictional force, refer here:

brainly.com/question/1714663

#SPJ1

Your question is incomplete. But your complete question probably was:

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?

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7 0

3 years ago

Read 2 more answers

A sock stuck to the inside of the clothes dryer spins around the drum once every 2.0 s at a distance of 0.50 m from the center o

Rashid [163]

a) 1.57 m/s

The sock spins once every 2.0 seconds, so its period is

T = 2.0 s

Therefore, the angular velocity of the sock is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s$

The linear speed of the sock is given by

$v=\omega r$

where

$\omega$ is the angular velocity

r = 0.50 m is the radius of the circular path of the sock

Substituting, we find:

$v=(3.14)(0.50)=1.57 m/s$

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

$r' = 2r = 1.00 m$

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

$\omega' = \omega = 3.14 rad/s$

Therefore, the new linear speed would be:

$v'=\omega' r' = \omega (2r)$

And substituting,

$v'=(3.14)(1.00)=3.14 rad/s = 2v$

So, we see that the linear speed has doubled.

8 0

3 years ago

Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th

alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box $m=3\ kg$

Force applied is $F=25\ N$

Displacement of the box is $s=15\ m$

Velocity acquired by the box is $v=6\ m/s$

acceleration associated with it is $a=\dfrac{F}{m}$

$\Rightarrow a=\dfrac{25}{3}\ m/s^2$

Work done by force is $W=F\cdot s$

$W=25\times 15\\W=375\ J$

change in kinetic energy is $\Delta K$

$\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J$

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

$\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J$

Therefore, the magnitude of work done by friction is $321\ J$

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3 years ago

driving down the highway , you find yourself behind a heavily loaded tomato truck. you follow close behind the truck keeping the

prohojiy [21]

Answer:

The tomato won't hit the car

Explanation:

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