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Likurg_2 [28]
3 years ago
11

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

Physics
1 answer:
krek1111 [17]3 years ago
3 0

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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A girl exerts a horizontal force of 109 N on a crate with a mass of 31.2 kg. HINT (a) If the crate doesn't move, what's the magn
vesna_86 [32]

Answer:

(a) Magnitude of static friction force is 109 N

(b) Minimum possible value of static friction is 0.356

Solution:

As per the question;

Horizontal force exerted  by the girl, F = 109 N

Mass of the crate, m = 31.2 kg

Now,

(a) To calculate the magnitude of static friction force:

Since, the crate is at rest, the forces on the crate are balanced and thus the horizontal force is equal to the frictional force, f:

F = f = 109 N

(b) The maximum possible force of friction between the floor and the crate is given by:

f_{m} = \mu_{s}N

where

N = Normal reaction = mg

Thus

f_{m} = \mu_{s}mg

For the crate to remain at rest, The force exerted on the crate must be less than or equal to the maximum force of friction.

f\leq f_{m}

f \leq \mu_{s}mg

109 \leq \mu_{s}\times 31.2\times 9.8

\mu_{s}\geq 0.356

7 0
2 years ago
If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 × 10^{-11} m^3/kgs^2 is the gravitational constant on Earth. M_1, M_2 are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}

\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}

\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

8 0
3 years ago
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