Answer to Question #104299 in Classical Mechanics for Miguel

Question

1. A force of 3.6 N acts towards the East on an object with mass m = 2.0 kg, while simultaneously a force of 2.2 N acts towards the South on the same object. What is its acceleration? ( correct number of sig fig).

a) Magnitude of the acceleration a = _____ m.s^-2 at

Using the information from question #703a above:
b) an angle of _____ ∘ South of East.

2. It is said that a mouse (a very small mammal) can fall 10 m and walk away, unharmed. If a moose (a very large mammal) fell 10 m it would probably not survive. Why is this? Select the best answer. (We don't say this often, but don't answer this one experimentally.)

a) the air resistance acting on the mouse is greater
b) the mouse decelerates over a shorter distance
c) the mouse's terminal speed is greater
d) the mouse hits the ground at a lower speed
e) the mouse is stronger than the moose

Answer 1

1.

As per the question,

Force towards east $F_e=3.6N$

Force towards south $F_s=2.2N$

mass of the object is m=2.0kg

As we know that south and east are perpendicular to each other, so the angle between these two force will be $90^\circ$

As per the parallelogram law of vector addition,

Net force on the object $(F)=\sqrt{F_e^2+F_s^2+2F_eF_s\cos 90}$

$\Rightarrow F=\sqrt{3.6^2+2.2^2}=\sqrt{12.96+4.84}=4.22N$

a) acceleration (a)$\dfrac{F}{m}=\dfrac{4.22}{2}=2.11m/sec^2$

b) Angle of resultant force =$\tan\theta=\dfrac{F_s}{F_e}=\dfrac{2.2}{3.6}=0.61$

$\Rightarrow \theta=\tan^{-1}(0.61)=31.38^\circ$ South- east

2)

When a mouse and the moose falls from the height 10m then mouse can be able to survive but moose is not able to survive, in the both two case there is two force working.

First is force of gravity and second is air resistance.

gravity force is vertically downwards and air resistance will be vertically upwards.

$F=mg-\dfrac{\rho A C V^2}{2}$

where m is the mass, g is the acceleration due to gravity, $\rho$ is the density of the object, A is the area of surface area of the object, C is the coefficient of friction, V is the terminal velocity. So when the object starts to fall, the force on the higher mass object will be high and the force on the lower mass object will be less.

The cross section of the mouse is less but the drag force will be high, it is because when the speed will get increase then due to increase, in speed, it will get in equilibrium because net force on it will get zero, (the gravitational force and the air resistance force balance), then the object will move at a constant speed—we call this the terminal velocity.

so b is the correct answer.