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3 years ago

6

A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a distance d. For the next loading, the spring is

compressed a distance 8d. How much work is required to load the second dart compared to that required to load the first?

1 answer:

Aliun [14]3 years ago

8 0

Answer:

the work required for the loading of second dart is 64 times greater as work required for loading the first dart.

Explanation:

k = spring constant of the spring loaded toy dart gun

x₁ = compression of spring to load the first dart = d

x₂ = compression of spring to load the second dart = 8 d

E₁ = Work required to load the first dart

E₂ = Work required to load the second dart

Work required to load the first dart is given as

E₁ = (0.5) k x₁² = (0.5) k d²

Work required to load the second dart is given as

E₂ = (0.5) k x₂² = (0.5) k (8d)² = (64) (0.5) k d²

E₂ = 64 E₁

So the work required for the loading of second dart is 64 times greater as work required for loading the first dart

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$t \to T$ --- time

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Hence:

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Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

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$v_0t = s$

Make $v_0$ the subject

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Replace s and t with their units

$v_0 = \frac{L}{T}$

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Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

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$a_0 = LT^{-2$ --- dimension

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Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

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$S_0 = LT^{-4$ --- dimension

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$s=\frac{ct^5}{120}$

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$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

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$c = LT^{-5}$

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$c \to LT^{-5}$ --- dimension

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