A block of mass 10, kg is kept on a rough inc | vedclass

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

Question

(A) Given: Mass $m = 10\, kg$,angle $	heta = 45^\circ$,coefficient of static friction $\mu = 0.6$,downward force $F_{down} = 3\, N$,$g = 10\, ms^{-2}

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(A) Given: Mass $m = 10\, kg$,angle $	heta = 45^\circ$,coefficient of static friction $\mu = 0.6$,downward force $F_{down} = 3\, N$,$g = 10\, ms^{-2}$.For the block not to move downward,the upward force $P$ must balance the downward components of gravity and the applied force,minus the maximum static friction force acting upwards.The downward force component due to gravity is $mg \sin 	heta = 10 	imes 10 	imes \sin 45^\circ = 100 	imes \frac{1}{\sqrt{2}} \approx 70.71\, N$.The normal force is $N = mg \cos 	heta = 10 	imes 10 	imes \cos 45^\circ = 100 	imes \frac{1}{\sqrt{2}} \approx 70.71\, N$.The maximum static friction force is $f_{max} = \mu N = 0.6 	imes 70.71 \approx 42.43\, N$.To prevent downward motion,the minimum upward force $P$ must satisfy: $P + f_{max} \geq mg \sin 	heta + 3\, N$.$P + 42.43 \geq 70.71 + 3$.$P \geq 73.71 - 42.43 = 31.28\, N$.Rounding to the nearest integer,the minimum force $P$ is $32\, N$.

Similar Questions

$A$ block is kept on an inclined plane of inclination $\theta$ and length $l$. The velocity of the block at the bottom of the inclined plane is (the coefficient of friction is $\mu$):

$A$ body is placed on a rough inclined plane of inclination $\theta$. As the angle $\theta$ is increased from $0^o$ to $90^o$,the contact force between the block and the plane

$A$ block of mass $m$ slides down a plane inclined at an angle $\theta$. Which of the following will $NOT$ increase the energy lost by the block due to friction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module