All categories

2 years ago

What is the minimum work needed to push a 950-kg car 710 m up along a 9.0Â° incline? ignore friction?

2 answers:

7 0

<span>1.0344645 MJ The minimum energy need is the potential energy of the car at the top of the ramp and is given by mass*gravity*height mass is known, gravity is assumed to be 9.81m/s^2 as it is on earth, and height must be calculated using trigonometry. height=sin(9 degrees)*710m=111meters so potential energy = 950kg*111m*9.81m/s^2=1.0344645 MJ Using the law of the conservation of energy we can assume that the energy expended to push the car up the incline was at least the potential energy gained by moving 111m against the pull of gravity.</span>

expeople1 [14]2 years ago

6 0

Answer:

Work done, $W=1.03\times 10^6\ J$

Explanation:

It is given that,

Mass of the car, m = 950 kg

Distance up to the incline, d = 710 m

Angle of inclination of the car is 9 degrees.

When the car is push in the inclined path, the vertical component of the force will act on it such that the work done is given by :

$W=mg\ sin\theta\times d$

$W=950\times 9.8\times \ sin(9) \times 710$

$W=1.03\times 10^6\ J$

So, the minimum work needed to push a car is $1.03\times 10^6\ J$ . Hence, this is the required solution.

You might be interested in

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

2 years ago

Part F - Example: Finding Two Forces (Part I)

Temka [501]

Answer: $F=28.936 kg/m s^{2}$

Explanation:

According to the given information (and figure attached), the block with mass $m=10 kg$ has the following forces acting on it:

In the X component:

$F cos(30\°) - F_{s}=0$ (1)

Where:

$F$ is the applied force directed $30\°$ above the horizontal

$F_{s}=\mu_{s} N$ (2) is the force of static friction (which is equal to the coefficient of static friction $\mu_{s}=0.3$ and the Normal force $N$

In the Y component:

$F sin(30\°) + N - W=0$ (3)

Where $W=m.g$ is the weight (the force of gravity) which is proportional to the multiplication of the mass $m$ and gravity $g=9.8 m/s^{2}$

Let’s begin by combining (1) and (2):

$F cos(30\°) - \mu_{s} N=0$ (4)

Isolating $N$ from (3):

$N=mg – F sin(30\°)$ (5)

Substituting (5) in (4):

$F cos(30\°) - \mu_{s} (mg – F sin(30\°))=0$ (6)

$F cos(30\°) - \mu_{s} mg + \mu_{s} F sin(30\°))=0$

$((cos(30\°) +\mu_{s} sin(30\°)) F - \mu_{s}mg =0$

Isolating $F$ :

$F=\frac{\mu_{s}mg}{(cos(30\°) +\mu_{s} sin(30\°)}$ (7)

$F=\frac{(0.3)(10 kg)(9.8 m/s^{2})}{(cos(30\°) + 0.3 sin(30\°)}$

Finally:

$F=28.936 N=8.936 kgm/s^{2}$ (8) This is the necessary force to overcome static friction and move the block

We can prove it by finding $F_{s}$ and verifying it is less than $F$ :

Substituting (8) in (1):

$8.936 kgm/s^{2}cos(30\°) - F_{s}=0$ (9)

$F_{s}=25.059 kgm/s^{2}$ (10) This is the static friction force

As we can see $F_{s} < F$

8 0

2 years ago

Some hydrogen gas is enclosed within a chamber being held at 200∘c with a volume of 0.0250 m3. the chamber is fitted with a mova

Mrac [35]

Answer: The final volume V₂ of the container is 0.039 m³.

Explanation:

Since the temperature is constant, the gas would expand isothermally.

For isothermal expansion,

P₁V₁=P₂V₂

Where, P₁ and P₂ are the initial and final pressure and V₁ and V₂ are initial and final volume.

It is given that:

V₁ = 0.0250 m³

P₁ = 1.5 × 10⁶ Pa

P₂ = 0.950 × 10⁶ Pa

V₂ = ?

⇒ 1.5 × 10⁶ Pa × 0.0250 m³ = 0.950 × 10⁶ Pa × V₂

⇒V₂ = 0.039 m³

Hence, the final volume V₂ of the container is 0.039 m³.

4 0

2 years ago

a student drops an object from the top of a building which is 19.6 m high. How long does it take the object to fall to the groun

zubka84 [21]

Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before.
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.

This formula tells how far an object travels in how much time,
when it's accelerating:

   Distance = (1/2 acceleration) x (Time²).

   D = 1/2 A T²

For your student who dropped an object out of the window,

   Distance = 19.6 m
   Acceleration = gravity = 9.8 m/s²

D = 1/2 G T²

    19.6 =   4.9   T²

Divide each side by 4.9 :    4 =   T²

Square root each side:    2 =   T

When an object is dropped in Earth gravity,
it takes 2 seconds to fall the first 19.6 meters.

8 0

2 years ago

Read 2 more answers

A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The

malfutka [58]

Answer:

$\theta = 23.32^o$