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3 years ago

A 2500-kg car is being pushed up a hill at an angle of 35 degrees. Determine the gravitational

Physics

1 answer:

velikii [3]3 years ago

3 0

Answer:

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

Explanation:

For this exercise we will use Newton's second law, let's set a reference system where the x axis is parallel to the plane, in the adjoint we can see the forces in the system.

sin 35 = Wₓ / W

cos 35 = W_y / W

Wₓ = W sin 35

W_y = W cos 35

Wₓ = 2500 9.8 sin 35

Wₓ = 14052.6 N

let's write the equations for each axis

and

Y axis

N-W_y = 0

N = W_y

X axis

F -Wₓ = m a

F = Wₓ + m a = mg sin 35 + m a

F = m (a + g sin 35)

let's calculate

F = 2500 (5 + 9.8 sin 35)

F = 26552.6 N

The force that pulls the car down is Wₓ = 14052.6 N and the one that pushes the car up is F = 26552.6 N

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Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

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3 years ago

What is the reaction force of the table with a weight of 558N

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Answer: reaction force = -558N

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2 years ago

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3 years ago

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Answer:

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Acceleration $a=-3.2m/sec^2$

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