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goblinko [34]
3 years ago
10

(i) a force of 35.0 n is required to start a 6.0-kg box moving across a horizontal concrete floor, (a) what is the coefficient o

f static friction between the box and the floor? (b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2. what is the coefficient of kinetic friction?
Physics
1 answer:
Serjik [45]3 years ago
4 0

a. The force applied would be equal to the frictional force.

F = us Fn

where, F = applied force = 35 N, us = coeff of static friction, Fn = normal force = weight

 

35 N = us * (6 kg * 9.81 m/s^2)

us = 0.595

 

b. The force applied would now be the sum of the frictional force and force due to acceleration

F = uk Fn + m a

where, uk = coeff of kinetic friction

 

35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)

uk = 0.533

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A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget
avanturin [10]

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

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using equation of motion

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now using equation

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7 0
3 years ago
A 3600-N Force causes a car to accelerate at a rate of 4m/s2. What is the mass of the car?
Mademuasel [1]

Answer:

14400 kg

Explanation:

According to Newton's second law of motion, F = ma

F = 3600 N

a = 4m/s2

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3600*4 = m

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Therefore the mass of the car is 14400 kg

Hope u understand

Please mark as brainliest

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