Question

if a block up an inclined plane 30 degree with a velocity of 5 m/s ,stops after 0.5 s then ; coffecient of friction will be ????????????????????????????????????

Answer 1

if a block up an inclined plane 30 degree with a velocity of $5\mathrm{m / s}$, stops after 0.5 s then; coefficient of friction will be ?

Solution:

$\mathrm{F_{fr}} -$ friction force;

$\alpha = 30{}^{\circ} -$ angle of the inclined plane;

$\mathrm{V}_0 = 5\frac{\mathrm{m}}{\mathrm{s}} -$ initial velocity;

$t = 0.5\mathrm{s} -$ time after the block stopped;

$a -$ deceleration of the block;

Newton's second law for the block on the X-axis:

$$x: F _ {f r} + m g _ {x} = m a$$

Formula for the friction force:

$$\mathrm {F} _ {\mathrm {f r}} = \mu \mathrm {N} \Rightarrow$$ $$\mu \mathrm {N} + \mathrm {m g} _ {\mathrm {x}} = \mathrm {m a}$$

Newton's second law for the block on the Y-axis:

$$y: N - m g _ {y} = 0$$ $$N = m g _ {y}$$

From the right triangle ABC:

$$\sin \alpha = \frac {m g _ {x}}{m g} \Rightarrow m g _ {x} = m g \sin \alpha$$ $$\cos \alpha = \frac {m g _ {y}}{m g} \Rightarrow m g _ {y} = m g \cos \alpha$$ $$N = m g _ {y} = m g \cos \alpha$$

(2)in(1):

$$\mu \mathrm {m g} \cos \alpha + \mathrm {m g} \sin \alpha = \mathrm {m a}$$ $$\mu \mathrm {g} \cos \alpha + \mathrm {g} \sin \alpha = \mathrm {a}$$

Rate equation of the block along X-axis:

$$z: 0 = V - a t$$ $$V = a t$$ $$a = \frac {V}{t}$$

(3)in(4):

$$\mu \mathrm {g} \cos \alpha + \mathrm {g} \sin \alpha = \frac {\mathrm {V}}{\mathrm {t}}$$ $$\mu = \frac {\frac {\mathrm {V}}{\mathrm {t}} - \mathrm {g} \sin \alpha}{\mathrm {g} \cos \alpha} = \frac {\mathrm {V}}{\mathrm {t} \cdot \mathrm {g} \cos \alpha} - \tan \alpha = \frac {5 \frac {\mathrm {m}}{\mathrm {s}}}{0. 5 \mathrm {s} \cdot 9 . 8 \frac {\mathrm {m}}{\mathrm {s} ^ {2}} \cdot \cos 3 0 {}^ {\circ}} - \tan 3 0 {}^ {\circ} = 0. 6$$

Answer: coefficient of the friction is $\mu = 0.6$