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6

Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes. For example, the work done on a mass $m$ lifted a distance $h$ against gravity with an acceleration $g$ is given by:$$W=F\times h=mgh$$ If you are told that someone is going to drop a $1$ kilogram mass on your head from a height of $10$ ...


5

Since the gravitational force only pulls the ball down, but not back or forth, it will not experience any acceleration changing its forward velocity but only downward acceleration. Thus, the ball will return to the thrower. You can also imagine the train to have no windows and be moving extremely smoothly. The thrower won't know if the train is moving or ...


5

$\int^t_0 A x^2 dt = x_0 + A x^2 t$ is incorrect. You are assuming $x$ as a constant. $x$ is a function of time x(t). Try $\dfrac{dx}{dt}=Ax^2 \implies \dfrac{dx}{x^2}=Adt$. Now integrate both the sides in appropriate limits. $$\int_{x_0}^{x(t)}\dfrac{dx}{x^2}=\int_0^t Adt$$ $$\int_{x_0}^{x(t)}x^{-2}dx=\int_0^t Adt$$ ...


4

Well, the reason it doesn't matter is that work is defined as $$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$ so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity. Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a ...


3

First problem: you say $v(t) = A x^2$, but that is a function of position, not time. Putting the definition right: $$ v = \frac{dx}{dt} = A x^2 $$ You can regroup terms on the same variable: $$ \frac{dx}{x^2} = A dt$$ And then do the integration: $$ \int \frac{dx}{x^2} = \int A dt$$ This is homework, so I will leave the integral limits and the ...


3

There is one formula relating the speeds of any two "platforms" (say $P$ and $Q$) between each other: $$V_{P}[ Q ] = V_{Q}[ P ].$$ And there's of course the well known symbol for "speed of light (in vacuum)", as determined of light signals exchanged by members of any one platform between each other: $c$. The speed of any one platform ($Q$) as determined ...


2

As you describe, the definition of work is just: $W=F d$. What you are confusing maybe is the rate of work $P$ and the force $F$. When you move fast, $P=Fv$ is larger, however the travelling time is shorter. let's consider we are moving in a constant velocity. Then: $$W=Pt=Fvt=Fd$$ Independent of velocity.


2

As you note, for a constant force acting on an object which moves in one direction, the work done is equal to $Fd$. One can see from the equation that work is not dependent on time, but only on force and displacement. In order to conceptualize this, you could think about the energy involved in the situation you describe. When work is applied by an external ...


2

Energy-momentum conservation is a stronger statement than the statement* that the inner product $p_\mu p'^\mu$ is conserved. It states that the sums are conserved individually/coordinatewise - $P_1+P_2=P_1'+P_2'$. As I see it, the conservation of the inner product is a statement about change in reference frames, whereas the conservation of energy and ...


2

The particle continues moving until it reaches the maximum height. At maximum height its velocity vanishes. In return, it again takes $T1$ to travel from $B$ to the ground. So it takes $\tfrac{T2 -T1}{2}$ for it to travel from B to the maximum height. $V_0$ denotes for initial velocity of particle on the ground and $H(B)$ the height of B, ...


2

Yes as answered above you are making mistake in approximating the values. You are taking speed to be constant for 1 sec. interval, rather it is changing everytime. Newton's equations are just based on integral method of using kinematic relations like $$s=\int v\,dt$$ $$v=\int a\,dt$$ So either trying using integral calculus or newton's equations.


2

Because the forces' magnitudes are equal, I would expect no net acceleration of $m$ Correct. Even if $m$ is moving, there will be no acceleration, since there is no net force. if $m$ is moving, the forces are doing work on $m$ ($F_1$'s work being the inverse of $F_2$'s work), whereas if $m$ is stationary, no work is done by either $F_1$ or $F_2$; ...


1

$$\Delta y = v_0 \sin(\theta) t - \frac12 gt^2$$ This is a projectile, so it will hit its max at $t_{max}=\frac12t$, where $t$ is the total time in which the projectile flies. The total time is when, as you know, $\Delta y=0$. Hence we've got: $$0=v_0 \sin(\theta) t - \frac12 gt^2$$ $$\require{cancel}\frac12 gt^\cancel{2}=v_0 \sin(\theta) \cancel{t}$$ ...


1

What may confuses you is the fact that there are two opposing forces doing work against each other if the body is moving and no forces at all if the body is at rest. This is a actually a nice example of the principle of relativity: Whether the body is at rest or whether it is moving with constant velocity shouldn't change the physics. :-) 'Two equal ...


1

The thing that is going wrong with your manual calculation is you are taking the velocity to be constant in every interval i.e., you are taking velocity to be $10m/s$ from $0$ to $1s$, $9m/s$ from $1s$ to $2s$ and so on which is incorrect. The velocity is continuously decreasing. You may calculate like this: At $t=0,v=10m/s,a=-1m/s^2$ which means from ...


1

The velocity after a time t1 of accelerating is the starting velocity of the deceleration phase. Thus $$a t_1 = b t_2$$ (not worrying about the sign here. I suppose you could) Further you have $$\frac12at_1^2+\frac12bt_2^2=s$$ Now you have two equations with two unknowns. Solving: rearrange first equation $$ \frac{a}{b} = \frac{t_2}{t_1}\\ t_2=\frac ab ...


1

Unfortunately I cannot comment due to insufficient reputation, so here a comment on the question. There are three cases: $\frac{1}{2}mv_A^2>2mgR$ In this case the pearl has a velocity $v>0$ in the top point and will continue its movement. $\frac{1}{2}mv_A^2<2mgR$ In this case the pearl won't reach the top and will oscillate around point $A$. ...


1

Work done is also defined as change in kinetic energy of the body. Since F is constant force so F/m=a is a constant acceleration of m. So, $$v^2-u^2=2ad$$or$$mv^2/2-mu^2/2=2mad/2$$which is the work done by the force. The body has travelled d distance with accleration a in the force field assuming u was a constant velocity when it entered the field and v is ...


1

Without any math and considering only Newtonian model here, I would say that if you move the inertial system at the same speed and direction as your mass point is moving, than you have no initial movement of the mass point and the total force used for acceleration will be the same as if you calculated or measured it in the original inertial system.


1

There are two formulas for adding velocities. The first is typically called Galilean relativity and the second special relativity. The first is simple, if you stack your tank on top of a train then the speed of the shell is the sum of the velocities, $v_1+v_2$. These things can be added as much as you like. You have an aircraft carrier moving at $v_1$ ...


1

See page 3 where $\Omega(t)$ is defined by: $$ \frac{dpos(t)}{dt} = \frac{\Omega(t) r}{K_r} $$ so the linear velocity of the vehicle in km/h is given by: $$ V(t) = \frac{3.6 r}{K_r} \Omega(t) \tag{1} $$ At the bottom of page 16 the constraint is introduced: $$ \Omega(t) \le \frac{K_r}{3.6 r} 50 $$ Comparing this with equation (1) shows that the number ...


1

If you were to draw out the velocity field, you would see that it is going outward. In fact if you were really astute, you would notice that your velocity has the same form as the electric field from a wire. This should help you visualize the field. Because of the symmetry, it makes sense to work in cylindrical coordinates. The position along the axis will ...


1

Your equations of motion are: $$ \Delta y = v_{0}\sin{\theta} \Delta t - \frac{1}{2}g\Delta t ^2 \\ \Delta x = v_{0}\cos{\theta}\Delta t $$ You can rearrange to eliminate $\Delta t$: $$ \Delta t = \frac{\Delta x}{v_0 \cos{\theta}} $$ Substituting this into the $\Delta y$ equation yields: $$ \Delta y = \Delta x \tan{\theta} - \frac{g(\Delta ...


1

Consider two masses M and m in circular motion with same velocity,v. Both has acceleration v^2/R. The forces acting on the two masses are different. Force will become more on the greater mass. But acceleration of both are same. Because, if you put M and m in the following relation, you get same v^2/R. $$(mv^2/R)/m=v^2/R$$ since we know $$F/m=a$$ where ...


1

You can think of acceleration from a purely mathematical context - it's the rate of change in velocity. (If you're familiar with calculus, you can say that the acceleration is the derivative of velocity.) Because of this, you don't need any mechanics to determine acceleration, so the mass is irrelevant. More concretely, a mathematician could calculate the ...


1

Mass and acceleration are two independent variables. You need to consider them together to arrive at a force with the direct relationship F=ma. In other words, if it's mass doubles, so does the force and the acceleration is the same. Gravity works exactly this way. If you are considering a fixed force from something other than gravity (a force of constant ...


1

I think you have too many parameters, and not all of the necessary ones To simplify your thinking: Change to a frame of reference in which one billiard ball is initially at rest at the origin, and the second is moving at velocity $V$ from right to left along the straight line $$y=k, \,k>=0$$ A collision will take place if and only if $k<2\sigma$. ...



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