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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 train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t
Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

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                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       V_{x} = mt + V_{o}      where m is the slope

Comparing equation (1) and (2)

V = V_{x}

a = m    

U = V_{o}

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        V_{x} = -1.40t + 22

                            V_{x} = -1.40(7.30) + 22

                             V_{x} = -10.22 + 22

                             V_{x} = 11. 78 m/s

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [\frac{1}{2} * (22 - 11.78) * (7.30)]  + [(11.78 - 0) * (7.30)]

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                           = 123. 297 m

                           ≈ 123. 30 m

                 

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