Physics, 02.11.2019 19:31 joeljuarez128oyusun

A6 kg block initially at rest is pulled to east along a horizontal, frictionless surface by a constant horizontal force of 12 n. find the speed of the block after it has moved 3m.

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Answer from: austinwilliams60

Answers:

a) 3.46 m/s

b) $2 m/s^{2}$

Explanation:

The complete question is written below:

A 6 kg block initially at rest is pulled to East along a horizontal, frictionless surface by a constant horizontal force of 12 N.

a) Find the speed of the block (using work-kinetic energy theorem) after it has moved 3.0 m.

b) Find the acceleration of the block using Newton.

a) Speed of the block after it has moved 3m

Let's begin by explaining the following:

The Work done by a Force refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path. In addition, is a scalar quantity, and its unit in the International System of Units is the Joule (like energy).

Now, when the applied force is constant and the direction of the force and the direction of motion are parallel, the equation to calculate it is:

W=F.d (1)

Where:

F= 12 N is the force exerted to pull the block to the East

d= 3 m is the distance the block has been moved

W=(12 N)(3 m) (2)

W=36 Nm= 36 J (3)

On the other hand, energy is the ability of matter to produce work in the form of movement, light, heat, among others. In this sense, according to the Work-Kinetic Energy Theorem we have:

$W=\Delta K=K_{f}-K_{o}$ (4)

Where:

$\Delta K$ is the variation in kinetic energy which is equal to:

$K=\frac{1}{2}mV^{2}$ (5)

Where:

m=6kg is the mass of the block

is the velocity

Hence, the Work-Kinetic Energy Theorem is written as follows:

$W=\frac{1}{2}mV_{f}^{2} - \frac{1}{2}mV_{o}^{2}$ (6)

Where the initial velocity is zero ( $V_{o}=0 m/s$ ) since we are told the block is initially at rest:

$W=\frac{1}{2}mV_{f}^{2}$ (7)

So, we have to find the final velocity $V_{f}$ , substituting (3) in (7):

$36 J=\frac{1}{2}6 kg V_{f}^{2}$ (8)

$V_{f}=3.46 m/s$ (9) This is the speed of the block after it has moved 3m

b) Acceleration of the block

According to Newton's second law of motion, which relates the force with the mass and the acceleration of an object, we have:

F=m.a (10)

On the other hand, acceleration is the variation of velocity $\Delta V$ in time $\Delta t$ :

$a=\frac{\Delta V}{\Delta t}$ (11)

Then, equation (10) is rewritten as:

$F=m \frac{\Delta V}{\Delta t}$ (12)

Isolating $\Delta t$ :

$\Delta t=\frac{m \Delta V}{F}$ (13)

Since $\Delta V=V_{f}-V_{o}$ and $V_{o}=0$ , we have:

$\Delta V=V_{f}=3.46 m/s$

$\Delta t=\frac{(6 kg) (3.46 m/s)}{12 N}$ (14)

$\Delta t=1.73 s$ (15)

Substituting (15) in (11):

$a=\frac{3.46 m/s}{1.73 s}$ (16)

Finally:

$a=1.99 m/s^{2} \approx 2 m/s^{2}$ This is the acceleration

Answer from: Quest

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

An object is located 5.0 cm from a convex lens. the lens focuses light at a distance of 10.0 cm. wha

Answer from: Quest

explanation:

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