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3 years ago

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What average force is required to stop a 1400 kg car in 6.0 s if the car is traveling at 90 km/h ? Express your answer to two si

gnificant figures and include the appropriate units. Enter positive value if the direction of the force is in the direction of the initial velocity and negative value if the direction of the force is in the direction opposite to the initial velocity.

1 answer:

devlian [24]3 years ago

8 0

Answer:

F=5833.3 N N

Explanation:

Newton's second law applied to the car

F= m*a Formula (1)

F: Force in Newtons (N)

m : mass in kg

a: acceleration ( m/s²)

kinematics car

vf= v₀ + a*t Formula (2)

vf : final velocity (m/s)

v₀ : final velocity (m/s)

a : acceleration ( m/s²)

t : time t

Equivalences

1 km= 1000m

1 h = 3600 s

Data

m= 1000kg

v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s

vf= 0

t= 6 s

Problem Development

We calculate the acceleration replacing the data in the formula (2) :

0 = 25 + a*6

a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)

We calculate the force is required to stop the car replacing the data in the formula (1)

-F = 1400 kg*(-4.16 m/s²)

F=5833.3 N

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a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

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d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

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Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

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b)

The period of the oscillating system is given by

$T=2\pi \sqrt{\frac{m}{k'}}$

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And for the system in the problem, we know that

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Solving the equation for m, we find the mass of the trailer:

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The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

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The period of the system is

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In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

$t=10T=10(2.09)=20.9 s$

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