# Thread: Physics Help

1. ## Physics Help

I have two questions for you guys. The first one should be simple yet the program used for our homework isn't accepting my answer. I'm really lost on the second question. Thanks for the help in advance.

First Question

In the figure below, a ball is shot directly upward from the ground with an initial speed of v0 = 7.60 m/s. Simultaneously, a construction elevator cab begins to move upward from the ground with a constant speed Vc = 4.00 m/s.

What maximum height does the ball reach relative to the floor of the elevator cab?

Using a kinematic equation, I found the maximum height the ball reaches which is 2.9469 m. The ball reaches its maximum height in .77551 s. I think I am misunderstanding the question, but should I just find the change in y for the elevator cab in .77551 s and subtract the change in y for the ball from the change in y for the cab?

Second Question

In the figure below, two blocks are connected over a pulley. The mass of block A is 15 kg, and the coefficient of kinetic friction between A and the incline is 0.22. Angle θ is 30°. Block A slides down the incline at constant speed. What is the mass of block B?

With constant speed, the forces acting on Block A should be 0. As a result, you should sum up the forces acting on the block.

I got for Block A: T-Wsin(x)+f=0

After this, I am completely lost. Inclined planes are not my strong suit.

Thanks for the help in advance.

2. ## Part 1

Yes, you are misunderstanding the first question. Let $y_b(t)$ be the height of the ball relative to the ground, and $y_e(t)$ the height of the floor of the elevator relative to the ground. The height of the ball relative to the floor of the elevator is $h=y_b-y_e$, and you want to find the maximum value of h (which will occur at a time earlier than the maximum of $y_b$ occurs). You should have that $y_b(t)=v_0t-\frac{1}{2}gt^2=7.60t-4.90t^2$ and that $y_b(t)=v_ct=4.00t$, so $h(t)=y_b-y_e=3.60t-4.90t^2$. To find the maximum, we find $h'(t)=3.60-9.80t$, set it equal to zero and solve for time, so that h is maximum at $t=\frac{3.60}{9.80}=0.367\,$ seconds, and you plug that into h(t) to get the maximum value of h.

--Kevin C.

3. ## Second problem

On the second problem,

Note that the component of A's weight perpendicular to the inclined plane is $W\cos\theta$, and so the normal force of the plane on A must cancel this, and is thus of magnitude $N=W\cos\theta$. Recalling that the magnitude of the force of friction is the magnitude of the normal force times the coefficient of friction, we get $f=\mu{N}=\mu{W}\cos\theta$, where $\mu$ is the coefficient of friction. So then our force equation for A becomes $T-W\sin\theta+\mu{W}\cos\theta=0$.

Now, look at the forces on B. As B will also be moving with constant velocity, the forces on it must cancel. Note that the forces on it are its weight downward and the tension upward. Thus $T=m_bg$, where $m_b$ is the mass of B, which is the unknown for which we are trying to solve. Thus we get:
$m_bg-W\sin\theta+\mu{W}\cos\theta=0$

Note that the weight of A is $W=m_ag$, where $m_a$ is the mass of A. Thus we can divide the force equation by g to get:
$m_b-m_a\sin\theta+\mu{m}_a\cos\theta=0$
$m_b=m_a\sin\theta-\mu{m}_a\cos\theta$
Now, just plug in the values for the mass of A (15 kg), the coefficient of friction (0.22), and the angle of incline (30°)

--Kevin C.

### a ball is shot directly upward from the ground

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