Kinetic Friction and Motion on Rough Horizontal Surface Questions in English

(A) The initial kinetic energy of the vehicle is given by $K = \frac{p^2}{2M}$.
When the engine is switched off,the only force acting to stop the vehicle is the kinetic friction force $f_k = \mu_k M g$.
The work done by the friction force in stopping the vehicle over a distance $s$ is equal to the change in kinetic energy.
$|W| = f_k \cdot s = \mu_k M g s$.
According to the work-energy theorem,the work done by friction equals the initial kinetic energy:
$\mu_k M g s = \frac{p^2}{2M}$.
Solving for $s$,we get:
$s = \frac{p^2}{2 M^2 \mu_k g}$.

(D) The block is moving on a rough horizontal surface. The forces acting on the block are the normal force $N$,gravitational force $mg$,the pulling force $F$,and the kinetic frictional force $fr_k$.
For vertical equilibrium,the normal force is $N = mg$.
The kinetic frictional force is given by $fr_k = \mu N = \mu mg$.
Power $P$ is defined as the product of the applied force $F$ and the velocity $v$ in the direction of motion: $P = F v$.
For the block to move at a constant velocity (which is its maximum velocity),the net force on the block must be zero.
Therefore,the pulling force $F$ must be equal to the kinetic frictional force $fr_k$:
$F = fr_k = \mu mg$.
Substituting this into the power equation:
$P = F v_{\max} = (\mu mg) v_{\max}$.
Solving for $v_{\max}$:
$v_{\max} = \frac{P}{\mu mg}$.

(B) Since the retardation is constant,the average velocity is given by:
$v_{av} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 \,m/s$
The frictional force $f$ is given by:
$f = \mu N = \mu mg$
Given:
Coefficient of friction,$\mu = 0.3$
Mass of body,$m = 500 \,g = 0.5 \,kg$
Acceleration due to gravity,$g = 10 \,m/s^2$
Substituting the values:
$f = 0.3 \times 0.5 \times 10 = 1.5 \,N$
The average power $P_{av}$ developed by the frictional force is:
$P_{av} = f \times v_{av}$
$P_{av} = 1.5 \,N \times 1 \,m/s = 1.5 \,W$
Thus,the absolute value of the average power is $1.5 \,W$.

(C) The boy is climbing the pole at a constant speed, which means his acceleration is $0$. Therefore, the net force acting on him in the vertical direction is $0$.
The forces acting on the boy in the vertical direction are his weight $(mg)$ acting downwards and the frictional force $(f)$ acting upwards.
For equilibrium, $f = mg$.
Given that the frictional force $f = \mu N$, where $\mu$ is the coefficient of friction and $N$ is the normal force (the horizontal force applied by the boy on the pole).
So, $\mu N = mg$.
Substituting the given values: $0.8 \times N = 40 \,kg \times 10 \,m/s^2$.
$0.8 \times N = 400 \,N$.
$N = \frac{400}{0.8} = 500 \,N$.
Thus, the horizontal force applied by the boy is $500 \,N$.