Question

A crate 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.

Answer 1

Answer:

(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 2

Answer:

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.

For more details, refer to the link:

https://brainly.com/question/17858145?referrer=searchResults