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Rina8888 [55]
4 years ago
13

Bruce (the dog) is pulling on his lead with a force of 40N at an angle of 26º below the horizontal. Show that the horizontal com

ponent of this force is about 36N.
Help???
Physics
2 answers:
saw5 [17]4 years ago
6 0
You will use the equation for work which is w= F times distance so you will take 40N times 36N which equals 1440N
Maslowich4 years ago
5 0

Answer:

The horizontal component of this force is about 36 N.

Explanation:

Given that,

Bruce is  is pulling on his lead with a force of 40 N

Angle below horizontal, \theta=26^{\circ}

We need to find the horizontal component of this force. The horizontal component of any vector is given by :

F_x=F\ \cos\theta

F_x=40\times \ \cos(26)

F_x=35.95\ N

or

F_x=36\ N

So, the horizontal component of this force is about 36 N. Hence, this is the required solution.

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Answer:

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Explanation:

Given:

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(a)

Using the equation of motion :

v^2=u^2+2a.d..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

u=0m.s^{-1}

putting the values in eq. (1)

v^2=0^2+2\times 3.62 \times 10

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Now, the momentum of the body just before the jump onto the tyre will be:

p=M.v

p=20\times 8.5088

p=170.1764\,kg.m.s^{-1}

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

(M+m)\times v'=p

(20+10)\times v'=170.1764

v'=5.6725\,m.s^{-1}

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

\frac{1}{2} (M+m).v'^2=(M+m).g.h

\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h

h\approx 1.6420\,m

above the normal hanging position.

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