A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firml
y in contact with the track as it goes around the loop. The magnitude of the block’s acceleration at the top of the loop isa) zero b) greater than zero but less than g c) equal to g d) greater than g
1 answer:
Answer:
d) greater than g
Explanation:
You have to apply Newton's Second Law to have an idea about the value of <em>the block's aceleration.</em>
∑F = ma
Where m is the mass, a is the acceleration and F represents all the forces involved.
You have to draw a free body diagram for the block at the top of the loop. (See the attachment)
Applying Newton's Second Law:
y: N+mg=ma
Where N is the <em>normal force</em> (the contact force that the loop's surface applies to the block) and mg is the <em>weight.</em>
Therefore a= + g
So, you can notice that <em>the magnitude of the block's acceleration</em> is greater than the acceleration of gravity because of the effect of the magnitude N/m.
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1) 148 J
When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:
where
mg is the weight of the object
is the change in height
For the box in this problem,
mg = 185 N
Substituting into the equation, we find:
2) (a) 28875 J
The work done by a force applied parallel to the direction of motion of the object is given by
where
F is the magnitude of the force
d is the displacement
In this problem,
F = 825 N is the force applied by the two students together
d = 35 m is the displacement of the car
Substituting,
2) (b) 57750 J
As seen previously, the equation that gives the work done by the force is
We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.
The work done previously was
W = 28875 J
Now the force is doubled, so the new work done will be
3) 4.4 J
In this case, the force acting on the ball is the force of gravity, whose magnitude is:
where
m = 0.180 kg is the mass of the ball
is the acceleration of gravity
Solving the equation,
Now we find the work done by gravity using the same formula applied before:
where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,
4) 595.2 kg
In this case, we have the work done on the box:
W = 7.0 kJ = 7000 J
And we also know the change in height of the box:
As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:
Solving for m, we find
And substituting the numerical values, we find the mass of the box:
5) They do the same work
In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:
Where is the difference in height between the final position and the initial position of the box.
This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of is the same for both of them, so the work they have done is exactly the same.
Answer:
See the explanation below
Explanation:
The watt (the power) is equal to the relationship between the work and the time in which that work is performed.
where:
W = work [J] (units of Joules)
t = time [s].
Now 1000 [W] are equal to 1 [kW]
And 1000000 [W] are equal to 1 [MW]
The horsepower is the unit of power in the imperial system of units.
And 745.7 [W] are equal to 1 [Hp]
Answer:
52.5°C
Explanation:
The final enthalpy is determined from energy balance where initial enthalpy and specific volume are obtained from A-12 for the given pressure and state
mh1 + W = mh2
h2 = h1 + W/m
h1 + Wα1/V1
242.9 kJ/kg + 2.35.0.11049kJ/ 0.35/60kg
=287.4 kJ/kg
From the final enthalpy and pressure the final temperature is obtained A-13 using interpolation
i.e T2 = T1 + T2 -T1/h2 -h1(h2 - h1)
= 50°C + 60 - 50/295.15 - 284.79
(287.4 - 284.79)°C
= 52.5°C
