Question

The capacitor plates are fixed on inclined plane p connected to a battery of emf E. A dielectric slab of mass m and dielectric constant k is inserted into the capacitor and tied to mass M using a weightless string as shown. If capacitance have plate area A, length l and distance d between them. Find M such that slab stays in equilibrium. Assume zero friction between slab and plates

Answer 1

Answer on Question 52652, Physics, Other

Question:

A $2kg$ box is projected with an initial speed of $3m/s$ up a rough plane inclined at $60{}^{\circ}$ above horizontal. The coefficient of kinetic friction is 0.3.

a) What is the energy dissipated by friction as the box slides up the plane?

b) What is the speed of the box when it again reaches its initial position?

Solution:

Let's write all forces that act on a box:

$$m \vec {g} + \vec {N} + \overrightarrow {F _ {f r}} = m \vec {a}$$

Then projected the forces on axis $x$ and $y$ :

$$- m g \sin \theta - F _ {f r} = m a,$$ $$N - m g c o s \theta = 0.$$

By the definition, the friction force is $F_{fr} = \mu_k N = \mu_k m g c o s \theta$ , and we can find the acceleration of the box from the first equation:

$$- m g \sin \theta - \mu_ {k} m g \cos \theta = m a,$$ $$a = - g (\sin \theta + \mu_ {k} \cos \theta).$$

a) The energy dissipated by friction as the box slides up the plane is equal to the work done on the box by the friction force as the box slides up the plane:

$$W_{fr} = F_{fr} s = \mu_k m g s c o s \theta,$$

where, $s$ is the distance, that the box slides up the plane before it stops momentarily (when $v = 0$).

We can find $s$ from the kinematic equation:

$$v^2 = v_0^2 + 2 a s.$$

Because $v = 0$ we get:

$$s = - \frac{v_0^2}{2 a} = \frac{v_0^2}{2 g (\sin \theta + \mu_k \cos \theta)} = \frac{\left(3 \frac{m}{s}\right)^2}{2 \cdot 9.8 \frac{m}{s^2} \cdot (\sin 60{}^\circ + 0.3 \cdot \cos 60{}^\circ)} = 0.452 m.$$

As we know $s$ we can find the energy dissipated by friction as the box slides up the plane:

$$W_{fr} = F_{fr} s = \mu_k m g s c o s \theta = 0.3 \cdot 2 k g \cdot 9.8 \frac{m}{s^2} \cdot 0.452 m \cdot \cos 60{}^\circ = 1.33 J.$$

b) In order to find the speed of the box when it again reaches its initial position we use the law of conservation of energy:

$$K E + P E = W_{g r a v. f o r c e} - W_{f r},$$

where, $KE$ and $PE$ is the kinetic energy and the potential energy of the box at initial position, $W_{grav.force}$ is the work done on the box by the gravitational force, $W_{fr}$ is the work done on the box by the friction force as the box slides down to its initial position, respectively.

Let's obtain the work done on the box by the gravitational force:

$$W_{g r a v. f o r c e} = \int_{s t a r t}^{e n d} \overrightarrow{F_g} \, d \vec{s},$$

where, $d\vec{s}$ is a vector along inclined rough plane in the direction of motion, and $\overrightarrow{F_g}$ is the gravitational force pointing down vertically.

So, we obtain:

$$\overrightarrow {F _ {g}} d \vec {s} = m g d s c o s \left(\frac {\pi}{2} + \theta\right) = - m g d s s i n \theta ,$$ $$W _ {g r a v. f o r c e} = \int_ {s} ^ {0} \overrightarrow {F _ {g}} d \vec {s} = \int_ {s} ^ {0} - m g s i n \theta d s = m g s s i n \theta .$$

Then, we can substitute $W_{grav.force}$ to the equation for the law of conservation of energy and obtain $v$:

$$\frac {1}{2} m v ^ {2} = m g s s i n \theta - \mu_ {k} m g s c o s \theta ,$$ $$\begin{array}{l} v = \sqrt {2 g s (s i n \theta - \mu_ {k} c o s \theta)} = \sqrt {2 \cdot 9 . 8 \frac {m}{s ^ {2}} \cdot 0 . 4 5 2 m \cdot (s i n 6 0 {}^ {\circ} - 0 . 3 \cdot c o s 6 0 {}^ {\circ})} = \\ = 2. 5 2 \frac {m}{s}. \\ \end{array}$$

Answer:

a) $W_{fr} = 1.33J$

b) $v = 2.52\frac{m}{s}$

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