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GREYUIT [131]
3 years ago
10

Calculate how much work you need to move the 130 N trunk to a ledge 2 m above.

Physics
2 answers:
kupik [55]3 years ago
8 0
Formula for the work is
W=Fs
where F is force and s is distance
Now we can just substitute our data and find work
W=130N*2m=260J
The result is 260J.
vaieri [72.5K]3 years ago
4 0
The absolute minimum work required for that job is the potential energy that will be added to the trunk by moving it up 2meters.

Potential energy = (weight) x (height) = (130 N) x (2) = <em>260 Joules</em>

That's the absolute minimum, because the trunk itself needs that much
energy in order to get up there.  But if you're doing the lifting, you may need
to expend more energy than that, just because the human body is not a
100% efficient machine.  It always takes more work to do something physical
than the energy you actually deliver to the job.

Here's a mind-blowing factoid: 
Do you see that 260 joules up there in the answer to the trunk ?
If you eat, let's say, 1800 Calories a day and don't gain any weight, that means
that you're burning off 260 joules of energy about every 3 seconds ! ! ! 
Just to keep your body warm, keep your breathing going, keep your brain
in tune, and move yourself around !  
Enough energy to lift that 30-pound trunk about 6-1/2 feet straight up off
the floor !


1 Calorie = 4184 joules
1 day = 86,400 seconds
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3 years ago
Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad
love history [14]

Answer:

a) The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}, b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}

Where:

a_{r} - Magnitude of the radial acceleration, measured in meters per square second.

a_{t} - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

\| \vec a \| = a_{t}

\| \vec a \| = r \cdot \alpha

Where:

\alpha - Angular acceleration, measured in radians per square second.

r - Radius of rotation (Radius of a tire), measured in meters.

Given that \alpha = 15\,\frac{rad}{s^{2}} and r = 0.31\,m. The linear acceleration experimented by the car is:

\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)

\| \vec a \| = 4.65\,\frac{m}{s^{2}}

The linear acceleration of the car is 4.65\,\frac{m}{s^{2}}.

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

Where:

\theta - Final angular position, measured in radians.

\theta_{o} - Initial angular position, measured in radians.

\omega_{o} - Initial angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

If \theta_{o} = 0\,rad, \omega_{o} = 0\,\frac{rad}{s}, \alpha = 15\,\frac{rad}{s^{2}}, the final angular position is:

\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}

\theta = 46.875\,rad

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

\theta = 7.46\,rev

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0
3 years ago
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