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3 years ago

Suppose an object’s initial velocity is 10 m/s and its final velocity is 4 m/s. Mass is constant.

Physics

1 answer:

galina1969 [7]3 years ago

8 0

The negative work done states that the work is done by the object and not on the object.

<u>Explanation: </u>

According to work energy theorem, the work done is equal to change in kinetic energy exhibited by the body. As the mass of the object is constant, and the velocity is decreased from 10 m/s to 4 m/s, the work done will be

$W=\frac{1}{2} \times\left(u^{2}-v^{2}\right)$

Here W is the work done, m is the mass and v is the final and u is the initial velocity of the object. As the initial velocity is greater than the final velocity. So

$W=\frac{1}{2} \times\left(4^{2}-10^{2}\right)=\frac{1}{2} \times(16-100)=\frac{1}{2} \times(-84)= - 42 J$

So the work done is negative for the given situation. The negative work done states that the work is done by the object and not on the object.

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A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?

Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

h = u² sin²θ/2g m

Where,

g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

h = 18² · sin²60 / 2 x 9.8

= 18² x 0.75 / 2 x 9.8

= 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0

4 years ago

A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first

Dominik [7]

Answer:

a) $V=14.01 m/s$

b) $V=8.86 \, m/s$

c) $t = 2.33s$

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

$E_m = E_k +E_p$

That is, mechanical energy is equal to the sum of potential and kinetic energy, and the value of this $E_m$ mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

$E_k = \frac{1}{2} m \, V^2$

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

$E_p= m\, g\, h$

where g is the acceleration due to gravity ( $g=9.81 \, m/s^2$ ) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which $h=0$ , a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

$E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m$

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set $h=0$

$E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2$

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

$E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\$

And so we have found the velocity of the ball as it hits the floor.

$V = \sqrt[]{2g\cdot 10m}=14.01\, m/s$

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

$V = \sqrt[]{2g\cdot h}$

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

$V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s$

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average? It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

$V_{avg}\cdot t=h\\t=h/V_{avg}$

what's the average speed when the ball is descending?

$V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s$

so the time it takes the ball to go down is:

$t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\$

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

$V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s$

and the time it takes to go up is: $t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s$

When we add both times , we get:

$t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s$

6 0

3 years ago

A jewellery shop owner has two identical clear gemstones but cannot remember which one is diamond and which one is rutile. Rutil

pantera1 [17]

We know

$\boxed{\sf n_{21}=\dfrac{C}{V}}$

How he find:-

The owner will measure speed of light through the index(Diamond or rutile).Then using calculations he fill find which is the velocity of diamond or rutile

<h3> For diamond</h3>

$\\ \sf\longmapsto n_D=\dfrac{C}{V_D}$