Physics, 04.12.2019 03:31 fonzocoronado3478

Acrate is pulled by a force (parallel to the
incline) up a rough incline. the crate has an
initial speed shown in the figure below. the
crate is pulled a distance of 8.05 m on the
incline by a 150 n force.
the acceleration of gravity is 9.8 m/s
2

a) what is the change in kinetic energy of
the crate?
answer in units of j.

b) what is the speed of the crate after it is
pulled the 8.05 m?
answer in units of m/s.

Answers: 3

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Answers

Answer from: AeelynRamos

The change in kinetic energy is 648.32 J.

The speed of crate after it pulled 8.05 m distance is 11.38 m/s.

Explanation:

Given data:

Distance covered by crate is, $d=8.05 \;\rm m$ .

Magnitude of force parallel to surface is, $F=150 \;\rm N$ .

Gravitational acceleration is, $g= 9.8 \;\rm m/s^{2}$

Angle of inclination is, $\theta =27^{\circ}$ .

Coefficient of friction is, $\mu = 0.286$ .

The mass of crate is, $m=10 \;\rm kg$

(a)

The change in kinetic energy is equal to work done on crate.

Change in kinetic energy = Work done

$KE =W\\KE= F \times d\\KE = m \times a \times d ...................................................... (1)$

Here, a is the acceleration.

Now, net force acting on block in an inclined plane is,

$F_{net}=F-\mu \times m \times gsin \theta\\m \times a=150-(0.286 \times 10 \times 9.8sin 27 ^\circ)\\10 \times a=150-0.286 \times 10 \times 9.8sin 27 ^\circ\\a=8.054 \;\rm m/s^{2}$

Substitute the value of a in equation (1) as,

$KE = 10 \times 8.054 \times 8.05\\KE = 648.32 \;\rm J$

Thus, the change in kinetic energy is 648.32 J.

(b)

Applying third kinematic equation of motion as,

$v^{2}=u^{2}+2ad\\v^{2}=0^{2}+2 \times 8.054 \times 8.05\\v =\sqrt{2 \times 8.054 \times 8.05} \\v =11.38 \;\rm m/s$

Thus, speed of block after covering 8.05 m distance is 11.38 m/s.

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Acrate is pulled by a force (parallel to the incline) up a rough incline. the crate has an initial s

Answer from: mireyagonzaless6395

(a) Kinetic energy of the cart is 648.32 J.

(b) Speed of the cart is 11.44 m/s.

Explanation:

From the attached below figure we came to know that,

Mass of the crate is 10 kg (m)

Initial speed of the crate is 1.2 m/s (v)

Coefficient of friction on a rough plane is 0.286 (µ)

Force on the cart is 150 N (F)

The distance which crate is moved (d) is 8.05 m.

$\text { Angle at which cart is moved is } 27^{\circ}(\theta)$

$\text { "g" acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}$

Calculation of kinetic energy:

Acceleration of the cart:

We know that,

F = F – m × g × sinθ - µ × m × g × cosθ (F = ma)

m × a = F – m × g × sinθ - µ × m × g × cosθ

10 × a = 150 – (10 × 9.8 × sin 27) – (0.286 × 10 × 9.8 × cos 27)

10 × a = 150 – 44.49 – 24.97

10 × a = 80.54

$\mathrm{a}=\frac{80.54}{10}$

$\mathrm{a}=8.054 \mathrm{m} / \mathrm{s}^{2}$

Change in kinetic energy = m × a × d

Change in kinetic energy = 10 × 8.054 × 8.05

Change in kinetic energy = 648.32 J

Calculation of speed:

$\text { Speed of the crate is }\left(v_{f}^{2}\right)=v^{2}+2 a d$

$\text { Speed of the crate is }\left(v_{\mathrm{r}}^{2}\right)=1.2^{2}+2 \times 8.054 \times 8.05$

$\text { Speed of the crate is }\left(v_{f}^{2}\right)=1.44+129.66$

$\text { Speed of the crate is }\left(v_{f}^{2}\right)=131.1$

$\text { Speed of the crate is }\left(\mathrm{v}_{\mathrm{f}}\right)=\sqrt{131.1}$

$\text { Speed of the crate is }\left(v_{f}\right)=11.44 \mathrm{m} / \mathrm{s}$

Answer from: Quest

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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Acrate is pulled by a force (parallel to the
incline) up a rough incline. the crate has an

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