Engineering College

2. The block is released from rest at the position shown, figure 1. The coefficient of
kinetic friction over length ab is 0.22, and over length bc is 0.16. Using the
principle of work and energy, find the velocity with which the block passes
position c.

2. The Block Is Released From Rest At The Position Shown, Figure 1. The Coefficient Ofkinetic Friction

Answers

Answer 1

Answer:

Velocity = 4.73 m/s.

Explanation:

Work done by friction is;

W_f = frictional force × displacement

So; W_f = Ff * Δs = (μF_n)*Δs

where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24

Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;

mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).

Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;

ΔKE2 + (0.16)(mg cos 24)(2).

Plugging in the relevant values, we have;

1.22mg = ΔKE1 + 0.603mg

ΔKE1 = 1.22mg - 0.603mg

ΔKE1 = 0.617mg

Also,

0.813mg = ΔKE2 + 0.292mg

ΔKE2 = 0.813mg - 0.292mg

ΔKE2 = 0.521mg

Now total increase in Kinetic Energy is ΔKE1 + ΔKE2

Thus,

Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg

Putting 9.81 for g to give;

Total increase in kinetic energy = 11.164m

Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m

m cancels out to give; ½v² = 11.164

v² = 2 × 11.164

v² = 22.328

v = √22.328

v = 4.73 m/s.

Related Questions

You want to plate a steel part having a surface area of 160 with a 0.002--thick layer of lead. The atomic mass of lead is 207.19 . The density of lead is 11.36 . How many atoms of lead are required

Answers

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

[tex] V = A*t [/tex]

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]    

Now, using the density we can find the mass:

[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]

Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):

[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline

Answers

Question:Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline . Which of them is correct ?

Answer: Technician  B is   correct

Explanation: Two types of engines exist , the two stroke (example, used in chainsaws)  is a type of engine that uses two strokes--a compression stroke and a return stroke to produce power in a crankshaft combustion cycle and the  four stroke engines(eg lawnmowers) which  uses four strokes,  2-strokes during  compression and exhaustion accompanied by 2 return strokes for each of the initial process to produce power in a combustion cycle.

While a 2 stroke system engine, requires mixing of oil and fuel to the crankshaft before  forcing  the mixture into the cylinder and do not require a pressurized system.  The 4 stroke system uses a splash and pressurized system where oil is not mixed with gasoline but drawn from the sump and  directed to  the main moving  parts of  crankshaft through its channels.

We can therefore say that Technician A is wrong while Technician B is  correct

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