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3 years ago

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The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

d dips in the road. When a 68 kg (about 150 lb) person sits on the left front fender of a small car, this corner of the car dips by about 1.2 cm (about 12 in). If we treat the spring assembly as a single spring, what is the approximate spring constant?

1 answer:

Marrrta [24]3 years ago

7 0

Answer: $k=55.590 KN/m$

Explanation:

Given

mass of person $\left ( m\right )$ =68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined spring constant

mg=kx

$k=\frac{mg}{x}$

$k=\frac{68\times 9.81}{1.2\time 10^{-2}}$

$k=55.590 KN/m$

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A 50.0-kg projectile is fired at an angle of 30 degrees above thehorizontal with an initial speed of 1.20 x 102 m/s fromthe top

Advocard [28]

Answer:

a) Em₀ = 42.96 104 J , b) $W_{fr}$ = -2.49 105 J , c) vf = 3.75 m / s

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The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has

Em = K + U

a) Let's look for the initial mechanical energy

Em₀ = K + U

Em₀ = ½ m v2 + mg and

Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142

Em₀ = 36 104 + 6.96 104

Em₀ = 42.96 104 J

b) The work of the friction force is equal to the change in the mechanical energy of the body

$W_{fr}$ = Em₂ -Em₀

Em₂ = K + U

Em₂ = ½ m v₂² + m g y₂

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Em₂ = 180.625 + 2.09 105

Em₂ = 1,806 105 J

$W_{fr}$ = Em₂ -Em₀

$W_{fr}$ = 1,806 105 - 4,296 105

$W_{fr}$ = -2.49 105 J

The negative sign indicates that the work that force and displacement have opposite directions

c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job

We have that the work of friction is equal to the change of mechanical energy

$W_{fr}$ = ΔEm

$W_{fr}$ = Emf - Emo

-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴

½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵

½ 50.0 vf² = 0.561

vf = √ 0.561 25

vf = 3.75 m / s

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