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vovangra [49]
2 years ago
9

A girl with mass m1 and a sled with mass m2 are on the frictionless ice of a frozen lake, a distance d apart but connected by a

rope of negligible mass. The girl exerts a horizontal force with magnitude F on the rope. What are the acceleration magnitudes of (a) the sled and (b) the girl? (c) How far from the girl's initial position do they meet? State your answers in terms of the given variables.
Physics
1 answer:
Charra [1.4K]2 years ago
6 0

Answer:

a)a_2=\dfrac{F}{m_2}

b)a_1=\dfrac{F}{m_1}

c)ex]x=\dfrac{m_2\times d}{m_1+m_2}[/tex]

Explanation:

Given that

Mass of girl =m_1

Mass of sled=m_2

Force = F

We know that from second law of Newton's

F =m a

a)

F=m_2.a_2

a_2=\dfrac{F}{m_2}

b)

F=m_1.a_1

a_1=\dfrac{F}{m_1}

c)

The distance ,x

x=\dfrac{m_1\times 0+m_2\times d}{m_1+m_2}

x=\dfrac{m_2\times d}{m_1+m_2}

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2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or
krek1111 [17]

a) The work done by the tree is -1.44\cdot 10^5 J

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

W=Fd

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

W=-1.44\cdot 10^5 J is the work done

d=50.0 cm = 0.50 m is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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The answer is B friction force
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Answer:

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Explanation:

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Step 1: Define

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