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Anna35 [415]
3 years ago
15

In our usual coordinate system( +x to the right, +y up (away from the center of the Earth), +z out of the page toward you), what

is the vector gravitational force on a 31 kg object sitting on the ground? (From the momentum principle you can conclude that a force of the same magnitude is exerted upward by the ground on the object if the momentum isn't changing. In Chapter 4 we'll see more about the origin of this force.)
Physics
1 answer:
skad [1K]3 years ago
5 0

Answer:

The vector gravitational force is  W = 303.8\  N( -y)

Explanation:

From the question we are told that

      The mass of the object is  m  = 31 \ kg

      The acceleration due to gravity experienced by the object is g =  9.8 m/s^2

Generally the vector gravitational force on the object is mathematically represented as

                       W = 31 *  9.8

                       W = 303.8\  N (-y)

The  -y indicates that the direction of the force is towards the center of the earth

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A horizontal spring-mass system has low friction, spring stiffness 165 N/m, and mass 0.6 kg. The system is released with an init
AURORKA [14]

a) 19.4 cm

b) 3.2 m/s

Explanation:

a)

A horizontal spring-mass system has a motion called simple harmonic motion, in which the mass oscillates following a periodic function (sine or cosine) around an equilibrium position.

As the system oscillates back and forth, its total mechanical energy (sum of elastic potential energy and kinetic energy) will remain conserved (since we consider friction negligible). The elastic potential energy at any point is given by:

U=\frac{1}{2}kx^2

where

k is the spring constant

x is the displacement of the system

While the kinetic energy at any point is

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

So the total mechanical energy of the system is

E=K+U=\frac{1}{2}mv^2+\frac{1}{2}kx^2

For this system, when it is initially released,

m = 0.6 kg

k = 165 N/m

x = 7 cm = 0.07 m

v = 3 m/s

So the total energy is

E=\frac{1}{2}(0.6)(3)^2+\frac{1}{2}(165)(0.07)^2=3.1 J

Since friction is negligible, this total energy remains constant. Therefore, when the system reaches its maximum stretch during the motion, the kinetic energy will be zero and all the mechanical energy will be elastic potential energy; so we will have:

E=U=\frac{1}{2}kx_{max}^2

where x_{max} is the maximum stretch. Solving for x_{max},

x_{max}=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(3.1)}{165}}=0.194 m

So, 19.4 cm.

b)

The maximum speed in a spring-mass oscillating system is reached when the kinetic energy is maximum, and therefore, since the total energy is conserved, when the elastic potential energy is zero:

U=0

which means when the displacement is zero:

x = 0

So, when the system is transiting through the equilibrium position.

Therefore, the total mechanical energy is equal to the maximum kinetic energy:

E=K=\frac{1}{2}mv_{max}^2

where

m is the mass

v_{max} is the maximum speed

Here we have:

E = 3.1 J

m = 0.6 kg

Therefore, solving for the maximum speed,

v_{max}=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(3.1)}{0.6}}=3.2 m/s

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