Starting from rest,a body slides down a 45^o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the distance be $s$ and the angle of inclination be $	heta = 45^o$.In the absence of friction,the acceleration is $a_1 = g \sin 	heta$. The

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(A) Let the distance be $s$ and the angle of inclination be $	heta = 45^o$.In the absence of friction,the acceleration is $a_1 = g \sin 	heta$. The time taken is $t_1 = t$.Using $s = \frac{1}{2} a_1 t_1^2$,we have $s = \frac{1}{2} (g \sin 	heta) t^2$.In the presence of friction,the acceleration is $a_2 = g(\sin 	heta - \mu \cos 	heta)$. The time taken is $t_2 = 2t$.Using $s = \frac{1}{2} a_2 t_2^2$,we have $s = \frac{1}{2} g(\sin 	heta - \mu \cos 	heta) (2t)^2 = 2 g(\sin 	heta - \mu \cos 	heta) t^2$.Equating the two expressions for $s$:$\frac{1}{2} g \sin 	heta t^2 = 2 g(\sin 	heta - \mu \cos 	heta) t^2$$\frac{1}{2} \sin 	heta = 2(\sin 	heta - \mu \cos 	heta)$$\sin 	heta = 4 \sin 	heta - 4 \mu \cos 	heta$$4 \mu \cos 	heta = 3 \sin 	heta$$\mu = \frac{3}{4} 	an 	heta$Since $	heta = 45^o$,$	an 45^o = 1$.Therefore,$\mu = \frac{3}{4} 	imes 1 = 0.75$.

Starting from rest,a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

Similar Questions

$A$ body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is $0.5$. The ratio of the net force required for the body to slide down and the normal reaction on the body is $1:2$. Then the angle of the inclined plane is (in $^{\circ}$)

The force required just to move a body up an inclined plane is double the force required just to prevent the body from sliding down. If the coefficient of friction is $0.25$,the angle of inclination of the plane is ...... $^o$.

The wedge is fixed and the mass of the block is $m$. The minimum value of $F$ so that the block is in equilibrium is (neglect the dimension of the block).

If the coefficient of friction between an ant and the surface is $1/3$,find the maximum value of $\alpha$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module