Question

A small rock with mass 0.28 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R = 0.46 m (Fig. E7.9). Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B at the bottom of the bowl has magnitude 0.22 J. (a) Between points A and B, how much work is done on the rock by (i) the normal force and (ii) gravity? (b) What is the speed of the rock as it reaches point B? (c) Of the three forces acting on the rock as it slides down the bowl, which (if any) are constant and which are not? Explain. (d) Just as the rock reaches point B, what is the normal force on it due to the bottom of the bowl?

Answer 1

Answer:

a.

$work \ due \ to \ normal \ force=0\\\\W_{grav}=1.2622J$

b.$v_2=3.0026\ m/s$

c. Normal Force-not constant

Frictional Force-not constant

Gravitational Force-constant

d. n=8.23N

Explanation:

#The mass of the rock m=0.20, the radius of the bowl , R=0.46m and the work done by friction between point A and B has a magnitude $W_f=0.22J$

=>The work total is expressed as:

$W_{tota}=W_{grav}+W_{other}=K_2-K_1\ \ \ \ \ \ \ ... i$

Work done by gravitational force is given as:

$W_{grav}=-\bigtriangleup U_{grav}=mgy_1+mgy_2\ \ \ \ \ \ ...ii$

And the kinetic energy is given by:

$K=\frac{1}{2}mv^2 \ \ \ \ \ \ \ ...iii$

a. The normal force is perpendicular to the displacement at time t of the motion, hence work done by the normal force is zero

ii. The potential energy is zero at point B, so we have:

$y_1=R=0.46, \ \ y_2=0\\\\W_{grav}=0.28\times 9.8\times 0.46-0\\\\=1.2622 J\\\\\therefore W_{grav}=1.26622J$

b. The speed of the rock as it reaches B can be calculated as:

The initial speed of the rock, $v_1=0$, and $\therefore K_1=0$.

#Since friction is acting opposite to the displacement, frictional work is -ve:

$W=-W_f=-0.22J$

#Substitute in eqtn 1:

$K_2=W_{grav}-W_f\\\\=1.2622-0.22=1.0422J\\\\\#substitute \ in \ iii\\\\1.0422=\frac{1}{2}\times 0.28.v_2^2\\\\v_2=\sqrt{1.0422/(0.5*0.28)}\\\\=3.0026\ m/s \\\\\\\therefore v_2=3.0026\ m/s$

c. A total of 3 forces are acting on the rock as it slides.

-Normal Force is dependent on the angle between the gravitational force and displacement. Gravitational force has a fixed direction  while the direction of displacement changes along its path , hence, the normal is not constant.

-Frictional force is given by $\mu_k n$ and is not constant.

-Gravitational Force,$F_g=mg$ is constant since $g=9.8ms^{-2}$ and the rock's mass are constants.

d. The rock's acceleration at point B is given as;

$a_{radial}=\frac{v^2}{R}$

=>Apply Newton's second law;

$\sum {F_y}=n-mg=ma_{radial}=m\frac{v^2}{R}\\\\\\n=m(\frac{v_2^2}{R}+g)$

We substitute the values in the equation;

$n=0.28[\frac{3.0026^2}{0.46}+9.8]\\\\=8.23 \ N$