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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
Physics
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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5.) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north.

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8 0
2 years ago
A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the
DiKsa [7]

Answer:

a). M = 20.392 kg

b). am = 0.56 m/s^2 (block),  aM = 0.28 m/s^2 (bucket)

Explanation:

a). We got  N = mg cos θ,

                  f = $\mu_s N$

                    = $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$   .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$  .............(iii)

   Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$   .....................(iv)

We got,   N = mg cos  θ

                $f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

  $T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$  ................(v)

Mg - 2T = Ma_M

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$    (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$   .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

a_M= 0.28 \ m/s^2

6 0
3 years ago
A car is driving around a banked curve, with the road surface at an angle of 10.0º. If the radius of curvature of the road is 30
IRISSAK [1]

Answer:

maximum speed 56 km/h

Explanation:

To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose

       sin 10 = Nx / N

      cos 10 = Ny / N

      Ny = N cos 10

     Nx = N sin 10

Let's develop Newton's equations on each axis

X axis

We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve

     Nx + fr = m a

     a = v2 / r

     fr = mu N

     N sin10 + mu N = m v² / r

     N (sin10 + mu) = m v² / r

Y Axis  

     Ny -W = 0

     N cos 10 = mg

Let's solve these two equations,

    (mg / cos 10) (sin 10 + mu) = m v² / r

    g (tan 10 + μ / cos 10) = v² / r

    v² = r g (tan 10 + μ / cos 10)

They ask us for the maximum speed

   v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)

   v² = 294 (0.8364)  

   v = √(245.9)

   v = 15.68 m / s

Let's reduce this to km / h

   v = 15.68 m / s (1 km / 1000m) (3600s / 1h)

   v = 56.45 km / h

This is the maximum speed so you don't skid

7 0
3 years ago
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