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lara [203]
3 years ago
13

A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet

ween the cup and the table is 0.30. Suddenly, the train accelerates. What is the maximum acceleration that the train can have without the cup sliding backward on the table?
Physics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

<u>a = 2.94 m/s²</u>

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a system of four particles moves along one dimension. the center of mass of the system is at rest, and the particles do not inte
Zinaida [17]

The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

The given parameters;

m1 = 1.45 kg, v1(t) = (6.09m/s) + (0.299m/s^2) × t

m2 = 2.81 kg, v2(t) = (7.83m/s) + (0.357m/s^2) × t

m3 = 3.89 kg, v3(t) = (8.09m/s) + (0.405m/s^2) × t

m4 = 5.03kg

The velocity of the center mass of the particles is calculated as;

McmVcm = m1v1 + m2v2 +m3v3+m4v4

Vcm= m1v1 + m2v2 +m3v3 +m4v4/ Mcm

0 = m1v1 + m2v2 +m3v3 +m4v4/ Mcm

m1v1 + m2v2 +m3v3+m4v4 = 0

m4v4 = -(m1v1 + m2v2 +m3v3)

v4 =-(m1v1 + m2v2 +m3v3)/ m4

The velocity of particle 1 at time, t = 2.83 s;

vi = 6.09 + 0.299× 2.83

v1 = 6.94 m/s

The velocity of particle 3 at time, t = 2.83 s;

v2 = 7.83 + 0.357 × 2.83

v2 = 8.84 m/s

The velocity of particle 3 at time, t = 2.83 s;

v3 = 8.09 + 0.405 × 2.83

v3 = 9.24 m/s

The velocity of particle 3 at time, t = 2.83 s;

v4 = - (m1v1 + m2v2 + m3v3)/m4

v4 = -(1.45×6.94 + 2.81×8.84 + 3.89×9.24)/5.03

v4 = -14.4 m/s

Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.

Learn more about Velocity here:

brainly.com/question/18084516

#SPJ4

5 0
1 year ago
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