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2 years ago

The value of the normal force exerted on a 2,000 gram block on a table

Physics

1 answer:

kobusy [5.1K]2 years ago

4 0

Answer:

N = 19.6 N

Explanation:

Given that,

Mass of a block, m = 2000 g

1 kg = 1000 g

It means, 2000 g = 2kg

We need to find the value of normal force on the block on a table. Normal force is balanced by the weight of the block as follows :

N = mg, g is acceleration due to gravity

N = 2 kg × 9.8 m/s²

N = 19.6 N

So, the normal force acting on the block is 19.6 N.

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2 years ago

The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of

Ahat [919]

Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is $=4.51m/s$

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

Kinetic energy of 9 kg block = initial energy stored - energy lost as a result of friction

Now to obtain the initial energy stored

Let U denote the initial energy stored and

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

Where x is the length the spring is displaced

k is the force constant of the string

$U = \frac{1}{2} * 627 * (0.6)^2$

$= 112.86 J$

Now referring to the formula above

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$\frac{1}{2} mv^2 = 112.86 - \mu_kmgx$

$v^2 = \frac{2(112.86 -\mu_kmgx)}{m}$

$v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}$

and we are told that coefficient of friction = 0.4 and the mass is 9 kg ,the acceleration due to gravity $= 9.8m/s^2$ this displacement length of spring = 0.6

Therefore $v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }$

$=4.51m/s$

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2 years ago

What is the velocity of a ball dropped from a height of 150 m when it hits the ground? Take the upward direction as positive.

djyliett [7]

Answer:

The velocity of the ball when its hit the ground will be 54.22 m/sec

Explanation:

We have given height from which ball is dropped h = 150 m

Acceleration due to gravity $g=9.8m/sec^2$

As the ball is dropped so initial velocity will be zero so u = 0 m/sec

According to third equation of motion we know that $v^2=u^2+2gh$

$v^2=0^2+2\times 9.8\times 150$

$v=54.22m/sec$

So the velocity of the ball when its hit the ground will be 54.22 m/sec

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3 years ago

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