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ahrayia [7]
3 years ago
7

A traveler pushes a suitcase 95.2 m with a 17.9 N force directed at 61.3° angle. How much work does she do? Unit =J

Physics
1 answer:
Sveta_85 [38]3 years ago
7 0
Work done (W) =  Horizontal force* distance moved

Horizontal force = F*Cos 61.3 = 17.9*Cos 61.3 = 8.596 N
Distance = 95.2 m

Therefore,

W= 8.596*95.2 = 818.34 J
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A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
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Answer:

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(iv) the resultant change would be constant.

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Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

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