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katrin2010 [14]
3 years ago
15

A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in

motion, a 127 N horizontal force keeps it moving at a constant velocity. Find the coefficient of static friction between the chair and the floor.
Physics
1 answer:
SCORPION-xisa [38]3 years ago
4 0

Answer:

\mu_k=0.51  

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction  = μk

The kinetic friction is given as follows

fk = μk  m g

Now by putting the values

127 = μk x 25 x 9.81

\mu_k=\dfrac{127}{25\times 9.81}

\mu_k=0.51

Therefore the value of coefficient of kinetic friction will be 0.51

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D. Mechanical energy remains unchanged.

Mechanical energy = Potential energy + Kinetic energy

So if potential energy decreases, then kinetic energy must increase so that the system is balanced.

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2 years ago
An attacker at the base of a castle wall 3.65 m high throws a rock straight up with speed 7.4m/s from a height of 1.55m above th
Natali5045456 [20]

a) Yes, the rock will reach the top

b) The final speed is 3.7 m/s

c) The change in speed is 2.4 m/s

d) The change in speed in the two situations do not agree

e) Because the kinetic energy depends quadratically on the speed, K\propto v^2

Explanation:

a)

The mechanical energy of the rock at the moment it is thrown from the ground is equal to the sum of its kinetic energy and its potential energy:

E=KE_i + PE_i = \frac{1}{2}mu^2 + mgh_i

where

m is the mass of the rock

u = 7.4 m/s is the inital speed

g=9.8 m/s^2 is the acceleration of gravity

h_i = 1.55 m is the initial height of the rock

Substituting, we find the initial mechanical energy of the rock

E=\frac{1}{2}m(7.4)^2 + m(9.8)(1.55)=42.6m [J]

In order to reach the top of the castle, the rock should have a mechanical energy of at least

E' = mgh'

where

h' = 3.65 m is the heigth of the top

Substituting,

E'=m(9.8)(3.65)=35.6m [J]

Since E > E', it means that the rock has enough mechanical energy to reach the top.

b)

The final mechanical energy of the rock at the top is

E=mgh'+ \frac{1}{2}mv^2 (1)

where:

v is the final speed of the rock at the top

Since the mechanical energy is conserved, this should be equal to the initial mechanical energy:

E=42.6 m [J] (2)

Therefore, equating (1) and (2), we can find the final speed of the rock:

mgh' + \frac{1}{2}mv^2 = 42.6m\\v=\sqrt{2(42.6-gh')}=\sqrt{2(42.6-(9.8)(3.65))}=3.7 m/s

c)

Since the motion of the rock is a free fall motion (constant acceleration equal to the acceleration of gravity), we can use the following suvat equation:

v^2 - u^2 = 2as

where

v is the final speed, at the bottom

u = 7.4 m/s is the initial speed of the rock, at the top

a=9.8 m/s^2 is the acceleration of gravity

s = 3.65 - 1.55 = 2.1 m is the vertical displacement of the rock

Solving for v, we find the final speed:

v=\sqrt{u^2+2as}=\sqrt{7.4^2 + 2(9.8)(2.1)}=9.8 m/s

Therefore, the change in speed is

\Delta v = v-u = 9.8 - 7.4 =2.4 m/s

d)

In the first situation (rock thrown upward), we have:

u = 7.4 m/s (initial speed)

v = 3.7 m/s (final speed)

So the change in speed is

\Delta v = v-u =3.7 - 7.4 = -3.7 m/s

While the change in speed in the second situation (rock thrown downward) is

\Delta v = 2.4 m/s

Therefore, we see that their magnitudes do not agree.

e)

In both situations, the change in kinetic energy of the rock is equal in magnitude to the change in gravitational potential energy, since the total mechanical energy is conserved.

The change in gravitational potential energy in the two situations is the same (because the change in height is the same), therefore the change in kinetic energy in the two situations is also the same.

However, the kinetic energy of the rock is not directly proportional to the speed, but to the square of the speed:

K\propto v^2

Since the initial speed is the same for both situation (7.4 m/s), but the change in kinetic energy has opposite sign in the two situations (negative when the rock is thrown upward, positive when thrown downward), the situation is not symmetrical, therefore in order to have the same magnitude of change in the kinetic energy, the change in speed must be larger when the kinetic energy involved is lower, so in the first situation.

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4 years ago
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matrenka [14]

B. A sandbar is formed by water. A sand dune is formed by wind.

4 0
3 years ago
The Burj Khalifa is the tallest building in the world at 828 m. How much work would a man with a weight of 700 N do if he climbe
8090 [49]

Answer:

579600J

Explanation:

Given parameters:

Height of the building  = 828m

Weight of the man  = 700N

Unknown:

Work done by the man  = ?

Solution:

The work done by the man is the same as the potential energy expended.

Work done:

            Work done  = Weight x height  = 700 x 828

       Work done  = 579600J

7 0
3 years ago
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