In the figure shown,all surfaces are friction | vedclass

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

Question

(C) For the block not to slip on the wedge,the pseudo force $ma$ must balance the component of gravity along the incline. The condition for no slippin

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(C) For the block not to slip on the wedge,the pseudo force $ma$ must balance the component of gravity along the incline. The condition for no slipping is $a = g 	an 	heta$. Given $a = 10\, m/s^2$ and $g = 10\, m/s^2$,we have $	an 	heta = 1$,so $	heta = 45^\circ$.The normal force $N$ acting on the block is $N = mg \cos 	heta + ma \sin 	heta$.Substituting the values: $N = (1)(10) \cos 45^\circ + (1)(10) \sin 45^\circ = 10(1/\sqrt{2}) + 10(1/\sqrt{2}) = 20/\sqrt{2} = 10\sqrt{2}\, N$.The block moves horizontally with the wedge with acceleration $a = 10\, m/s^2$. The displacement $s$ in time $t = \sqrt{3}\, s$ is $s = \frac{1}{2} a t^2 = \frac{1}{2} 	imes 10 	imes (\sqrt{3})^2 = 5 	imes 3 = 15\, m$.The normal force $N$ acts at an angle $	heta = 45^\circ$ with the vertical,so its horizontal component is $N_x = N \sin 	heta = (10\sqrt{2}) \sin 45^\circ = 10\sqrt{2} 	imes (1/\sqrt{2}) = 10\, N$.The work done by the normal force is $W = N_x 	imes s = 10\, N 	imes 15\, m = 150\, J$.

Similar Questions

$A$ force of $5\,N$ acts on a body of mass $5\,kg$ for $1\,s$ and gives it a momentum $p$ and kinetic energy $E$. If the same force accelerates the same body through $1\,m$,the momentum and energy attained by the body are $p'$ and $E'$ respectively. Which of the following relations is correct?

Three forces $\vec{F}_1=(2 \hat{i}+4 \hat{j}) \,N$,$\vec{F}_2=(2 \hat{j}-\hat{k}) \,N$,and $\vec{F}_3=(\hat{k}-4 \hat{i}-2 \hat{j}) \,N$ are applied on an object of mass $1 \,kg$ at rest at the origin. The position of the object at $t=2 \,s$ will be:

$10,000$ small balls,each weighing $1 \, g$,strike $1 \, cm^2$ of area per second with a velocity of $100 \, m/s$ in a normal direction and rebound with the same velocity. The value of pressure on the surface will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module