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Setler [38]
3 years ago
9

A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

t is the value of c if the work done by on the particle is (a) zero, (b) 4.0 J, and (c) -1.8 J
Physics
1 answer:
allsm [11]3 years ago
7 0

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

a)

The displacement of the particle is given by

d=2.2i+cj

While the force applied on the particle is

F=3.2i-3.8 j

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

W=F\cdot d (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85

b)

In this problem, the work done on the particle is

W=4.0 J

The force and displacement are still

d=2.2i+cj (displacement)

F=3.2i-3.8 j (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

W=F\cdot d

4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8

c)

In this problem instead, the work done on the particle is negative:

W=-1.8 J

As before, the force and displacement are

d=2.2i+cj (displacement)

F=3.2i-3.8 j (force)

And so again, we calculate the scalar product between  force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

W=F\cdot d

-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33

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a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
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Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

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        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

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we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

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         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

8 0
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