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100 POINTS FOR CORRECT ANSWER/EXPLANATION

Shalnov [3]

Answer:

6 N

Explanation:

Let's start with the small block m on top. There are four forces:

Weight force mg pulling down, normal force N₁ pushing up, tension force T pulling right, and friction force N₁μ pushing left (opposing the direction of motion).

Now let's look at the large block M on bottom. There are seven forces:

Normal force N₁ pushing down (opposite and equal from block m),

Friction force N₁μ pushing right (opposite and equal from block m),

Weight force Mg pulling down,

Tension force T pulling right,

Applied force F pulling left,

Normal force N₂ pushing up,

and friction force N₂μ pushing right (opposing the direction of motion).

So you've correctly identified the free body diagrams.

Now apply Newton's second law. Sum of forces in the y direction for block m:

∑F = ma

N₁ − mg = 0

N₁ = mg

Sum of forces in the x direction:

∑F = ma

T − N₁μ = 0

T = N₁μ

T = mgμ

Sum of forces in the y direction for block M:

∑F = ma

-N₁ − Mg + N₂ = 0

N₂ = N₁ + Mg

N₂ = mg + Mg

Sum of forces in the x direction:

∑F = ma

N₁μ + T − F + N₂μ = 0

F = N₁μ + T + N₂μ

F = mgμ + mgμ + (mg + Mg)μ

F = gμ(3m + M)

Since M = 2m:

F = 5gμm

Plug in values:

F = 5 (10 m/s²) (0.400) (0.300 kg)

F = 6 N

8 0

3 years ago

Read 2 more answers

A gas undergoes a process in a piston–cylinder assembly during which the pressure-specific volume relation is pv1.3 = constant.

Galina-37 [17]

Answer:

Change in specific internal Energy $=250\ \rm Btu/lb$

Explanation:

Given:

Mass of the gas, m=0.4 lb
Initial pressure and volume are $p_1=160\ \rm lbf/in^2\ and\ v_1=1\ \rm ft^3\\$

Final pressure and temperature are $p_1=480\ \rm lbf/in^2$
Heat transfer from the gas is 2.1 Btu

Since the process is isotropic we have

$p_1v_1^{1.3}=p_2v_2^{1.3}\\160\times 1^{1.3}=480\times v_2^{1.3}\\v_2=0.43\ \rm ft^3\\$

So the final volume of the gas is calculated.

Work in any isotropic is given by w

$w=\dfrac{p_1v_1-P_2v_2}{n-1}\\\\w=\dfrac{160\times1-480\times0.43}{1.3-1}\\w=-154.67\ \rm Btu\\$

According to the first law of thermodynamics we have

$Q=\Delta U+w\\-2.1=\Delta U-154.67\\\Delta U=152.56\ \rm Btu\\$

So the Specific Internal Change is given by

$\Delta u=\dfrac{\Delta U}{m}\\\Delta u=\dfrac{152.56}{0.4}\\\Delta u=250\ \rm Btu/lb$

So the specific Change in Internal energy is calculated.

6 0

3 years ago

A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:

olya-2409 [2.1K]

To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.

PART A ) The Impulse can be calculcated as follows

$L= F\Delta t$