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jok3333 [9.3K]
3 years ago
6

The velocity of an object with mass = 2kg is given as a function of time:

Physics
1 answer:
Musya8 [376]3 years ago
8 0

Answer:

The force acting on the object at t = 2\,s is \vec F = (4, 32)\,[N].

Explanation:

Given that object has a constant mass in time, the force acting on the object (\vec F), in newtons, is defined by following expression:

\vec F = m\cdot \vec a (1)

Where:

m - Mass, in kilograms.

\vec a - Acceleration, in meters per square second.

By definition of acceleration, we know that:

\vec a = \frac{d}{dt} \vec v (2)

Let suppose that given vector velocity is expressed in meters per second. If we know that m = 2\,kg, \vec v = (2\cdot t, 4\cdot t^{2})\,\left[\frac{m}{s} \right] and t = 2\,s, then the force acting on the object is:

\vec a = (2, 8\cdot t)\,\left[\frac{m}{s^{2}} \right]

\vec F = (4, 16\cdot t)\,[N]

\vec F = (4, 32)\,[N]

The force acting on the object at t = 2\,s is \vec F = (4, 32)\,[N].

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Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

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So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

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4 years ago
Two objects are moving at equal speed along a level, frictionless surface. The second object has twice the mass of the first obj
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Answer:

E)The two objects rise to the same height : h₁=h₂ =( v²) / (2g)

Explanation:

With  coefficient of kinetic friction,  μk =0:

We apply the principle of energy conservation :

E₀ = Ef Formula (1)

K₀+U₀ = Kf + Uf Formula (2)

K = (1/2) *m*v² Formula (3)

U = m*g*h Formula (4)

Where:

E₀:  

Initial total energy (J)

Ef:   Final total energy  (J)

K₀:   Initial kinetic energy (J)

U₀:  Initial potential energy (J)

Kf:  Final kinetic energy (J)

Uf:

Final kinetic energy (J)

v : speed (m/s)

m: mass (kg)

h : hight (m)

Known data

v₁=v₂=v

m₁=m

m₂ =2m

μk =0 : coefficient of kinetic friction

Problem development

We apply formulas (1), (2), (3) and (4)  for the  first object ,(1):

E₀₁ = Ef₁

K₀₁+U₀₁ = Kf₁ + Uf₁  U₀₁=0 ,  Kf₁ =0 , m₁=m

(1/2) *m*v²+0 =0+m*g*h₁    :We eliminate m,then,

(1/2) *v²=g*h₁

h₁ =( v²) /(2g)

We apply formulas (1), (2), (3) and (4)  for the  second object ,(2):

E₀₂ = Ef₂

K₀₂+U₀₂ = Kf₂ + Uf₂  U₀₂=0 ,  Kf₂ =0 , m₂=2m

(1/2) *2m*v²+0 =0+2m*g*h₂    :We eliminate m,then,

(1/2) *2*v² =2*g*h₂  : We divide by 2 on both sides of the equation,then,

(1/2) *v²=g*h₂

h₂ =( v²) / (2g)

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