Ask question

Ask question

All categories

4 years ago

5

A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j

is vertical if the box moves 10m horizontal in 90 seconds what is the work and power that is required to complete the movement

1 answer:

tamaranim1 [39]4 years ago

5 0

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

<u>Work = (force) x (distance)</u>

Work = (53.91 N) x (10 m)

<em>Work = 539.1 Joules</em>

<u>Power = (work done) / (time to do the work)</u>

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

<em>Power = 5.99 watts</em>

=====================================

Note: If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds. I get 109,168 meters !

You might be interested in

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

4 years ago

Simon is riding a bike at 12 km/h away from his friend Keesha. He throws a ball at 5 km/h back to Keesha, who is standing still

pychu [463]

So there are different ways this could be solved. I'll do try to explain it the way I was taught...

Simon is riding his bike at 12 km/hr relative to the sidewalk, away from where Keesha is.

Simon throws the ball at Keesha, at 5 km/hr.

Keesha sees the ball approaching her at (12-5) = 7 km/hr relative to the ground to her.

Therefore the answer is: 7 km/hr

3 0

3 years ago

An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) dur

Lilit [14]

Answer:

The thermal efficiency is 0.4113.

Explanation:

We know that the thermal efficiency is the ratio of work done by the engine over the heat taken

$\eta = \frac{W_{made}}{Q_{in}}$

Now, how much work the engine do in a cycle?

We know that the work done in a cycle must be equal to the heat taken minus the heat rejected

$W{made} = Q_{in} - Q_{rejected}$

So, the thermal efficiency will be:

$\eta = \frac{Q_{in} - Q_{rejected}}{Q_{in}}$

$\eta = \frac{Q_{in}}{Q_{in}} - \frac{Q_{rejected}}{Q_{in}}$

$\eta = 1 - \frac{Q_{rejected}}{Q_{in}}$

Putting the values of the problem

$\eta = 1 - \frac{259.8 kJ }{441.3kJ}$

$\eta = 0.4113$

7 0

4 years ago

The negative work done is equal to the difference in the......... Electric potential energy Electric flux Electric potential Ele

Leto [7]

Answer:

Electric potential energy

Explanation:

The Electric potential energy of a system is less than that carried out by electrostatic forces during the development of the system (as long as charges are initially cut infinitely).

The change in potential energy between an initial and final configuration is equal to minus the work done by the electrostatic forces.

8 0

3 years ago

1. What is the momentum of a 0.15 kg arrow that is traveling at 120 m/s?

Korvikt [17]

For this case we have that by definition, the momentum equation is given by:

$p = m * v$

Where:

m: It is the mass

v: It is the velocity

According to the data we have:

$m = 0.15 \ kg\\v = 120 \frac {m} {s}$

Substituting:

$p = 0.15 * 120\\p = 18 \frac {kg * m} {s}$

On the other hand, if we clear the variable "mass" we have:

$m = \frac {p} {v}$

According to the data we have:

$p = 250 \frac {kg * m} {s}\\v = 5 \frac {m} {s}\\m = \frac {250} {5}\\m = 50 \ kg$

Thus, the mass is $50 \ kg$

Answer:

$p = 18 \frac {kg * m} {s}\\m = 50 \ kg$

3 0

3 years ago