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3 years ago

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A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin

g this will be enough initial speed that he can make it all the way down this hill and up the next hill which is 45 m tall. Neglecting friction and air resistance, what does the initial speed need to be for him to make it up this second hill?

1 answer:

Strike441 [17]3 years ago

7 0

Answer:

This question can be answered by using conversation of energy.

$K_1 + U_1 = K_2 + U_2$

$\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2$

$\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s$

Explanation:

Note that we take $K_2 = 0$ because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.

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Read 2 more answers

A friction force of 280 N exists between a cart and the path. If the force of reaction is 766 N, what is the minimum action forc

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force of friction on the cart is given as

$F_f = 280 N$

here we also know the reaction force due to surface

$R = 766 N$

so we can say reaction force is given as

$R = \sqrt{F_n^2 + F_f^2}$

$766 = \sqrt{F_n^2 + 280^2}$

$F_n^2 = 766^2 - 280^2$

$F_n = 713 N$

now by force balance we will say

$F_y + F_n = mg$

$F_y = mg - F_n$

$F_x = F_f$

also we know that

$F_f = \mu * F_n$

$280 = \mu * 713$

$\mu = 0.39$

now minimum force required to set this into motion

$F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$

here we know that

$mg = F_n = 713 N$

$F_{min} = \frac{0.39* 713}{\sqrt{1 + 0.39^2}}$

$F_{min} = 259 N$

So it will require 259 N minimum force to move it

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3 years ago