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Lostsunrise [7]
3 years ago
8

A student librarian lifts a 2.2 kg book from the floor to a height of 1.25 m. He carries the book 8.0 m to the stacks and places

the book on a shelf that is.35 m above the floor. How much work does he do on the book?
I understand most of it, but I am unsure of what distance to use. Please help? ...?
Physics
2 answers:
Montano1993 [528]3 years ago
5 0
The solution to your problem is as follows:


2.2Kg*9.8m/s = 21.56N 
<span>
21.56N*1.25m = 26.95J </span>


<span>We're only concerned with the work done against gravity, lifting the books to 1.25 meters. the distance walked has no effect on the problem, unless you take into account the wind resistance and the force needed to overcome it. Also, lowering the books onto the shelf doesnt count, because gravity does the work on the books.</span>
german3 years ago
5 0

Answer:

W_{tot}=W_{1}+W_{2}=mgh_{2}=7.55[J]

Explanation:

The work done on an object is the scalar product between force and displacement. So we can write the work:

W=F\cdot d=Fd

In our case F and d are parallels then we have a common product of their magnitudes.

Now, the total work will be the sum these two works:

  1. Work done when the student librarian lifts a 2.2 kg book from the floor to a height of h₁=1.25 m.
  2. Work done when he places the book on a shelf that is h₂=0.35 m above the floor.

Let's recall that the force in this problem is just the weight of the book. F=m*g

  • The first work will be: W_{1}=mgh_{1}. F and h1 are parallels
  • The second work will be:W_{2}=-mg(h_{1}-h_{2}) is negative because the vector force and the vector displacement are anti parallels.

Finally, the total work will be the sum of W₁ and W₂.

W_{tot}=W_{1}+W_{2}=mgh_{2}=2.2*9.81*0.35=7.55[J]

I hope it helps you!

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3 years ago
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In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

ii) The force on friction acting on the blocks is proportional to their mass, since mass of block B is less than block A, the force of friction acting on block B is also less. Hence, one might argue that block B will go farther along the table before coming to rest.

B) The equation of motion for block A is

m_{A}\frac{\mathrm{d} v}{\mathrm{d} t} = -m_{A}g\nu_{s}\Rightarrow \frac{\mathrm{d} v}{\mathrm{d} t} = -\nu_{s}g \quad (1)

Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

v(t) = v_{0} - \nu_{s}gt \quad (4)

Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

Here, x(t) is the distance travelled by the block and D is again a constant of integration which can be determined by imposing the initial condition

x(t=0) = 0\Rightarrow D = 0 \quad (7)

The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

C) We can see that the expression for the distance travelled for block A is independent of its mass, therefore if we do the calculation for block B we will get the same result. Hence the reasoning for Student A and Student B are both correct, the effect of having larger initial energy due to larger mass is cancelled out by the effect of larger frictional force due to larger mass.

D)

i) The block A is moving in a circle of radius L+\frac{d}{2} , centered at the pivot, this is the distance of pivot from the center of mass of the block (assuming the block has uniform mass density). Because of circular motion there must be a centripetal force acting on the block in the radial direction, that must be provided by the tension in the string. Hence

T = \frac{m_{A}v^{2}}{L+\frac{d}{2}} \quad (9)

The speed of the block decreases with time due to friction, hence the speed of the block is maximum at the beginning of the motion, therfore the maximum tension is

T_{max} = \frac{m_{A}v_{0}^{2}}{L+\frac{d}{2}} \quad (10)

ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
3 years ago
A 2.0 kg block has a rope attached to the block on a table and is pulled with a force of 8.0 N. The block accelerated at 2.5m/s^
Masja [62]

Answer:

0.15

Explanation:

Assuming the rope is horizontal, sum the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum the forces in the x direction:

∑F = ma

F − Nμ = ma

Substitute:

F − mgμ = ma

mgμ = F − ma

μ = (F − ma) / (mg)

Plug in values:

μ = (8.0 N − 2.0 kg × 2.5 m/s²) / (2.0 kg × 9.8 m/s²)

μ = 0.15

3 0
2 years ago
The mass of jupiter is 300 times the mass of the earth. Jupiter orbits the sun with Tjupiter = 11.9 yr in an orbit with Rjupiter
mihalych1998 [28]

Answer:

c) 11.9 yr

Explanation:

The orbital period is proportional to r^(3/2) and does not depend on the satellite's mass. Any object at Jupiter position will have the same orbital period regardless of mass.

By keppler's law  we know that

T^2= r^3

T= orbital time period

r= mean distance of the planet from the Sun.

clearly, The orbital period does not depend on the satellite's mass

there, the correct answer will be c= 11.9 yr.

5 0
3 years ago
A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica
grin007 [14]

Answer:

f_e = 1.51 cm

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}

400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}

9.6 = \dfrac{16-f_e}{f_e}

9.6f_e = 16-f_e

10.6 f_e = 16

f_e = 1.51 cm

3 0
3 years ago
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