An engine of mass 1 metric ton is ascending a | vedclass

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

Question

(A) Given: Mass $m = 1000 \; kg$,speed $v = 36 \; km/h = 36 	imes (5/18) = 10 \; m/s$,$	an 	heta = 1/2$,$\mu = 1/\sqrt{3}$.From the triangle,$\sin

Vedclass · Accepted Answer

$94.4 	imes 10^3 \; W$

Vedclass · Answer

(A) Given: Mass $m = 1000 \; kg$,speed $v = 36 \; km/h = 36 	imes (5/18) = 10 \; m/s$,$	an 	heta = 1/2$,$\mu = 1/\sqrt{3}$.From the triangle,$\sin 	heta = 1/\sqrt{5}$ and $\cos 	heta = 2/\sqrt{5}$.The force required to move the engine up the incline at a constant speed is $F = mg \sin 	heta + f_k = mg \sin 	heta + \mu mg \cos 	heta$.$F = mg(\sin 	heta + \mu \cos 	heta) = 1000 	imes 9.8 	imes (1/\sqrt{5} + (1/\sqrt{3}) 	imes (2/\sqrt{5}))$.$F = 9800 	imes (1/\sqrt{5} + 2/\sqrt{15}) = 9800 	imes (0.4472 + 0.5164) \approx 9800 	imes 0.9636 \approx 9443 \; N$.Power $P = F 	imes v = 9443 	imes 10 = 94430 \; W = 94.4 	imes 10^3 \; W$.

An engine of mass $1$ metric ton is ascending an inclined plane,at an angle $\theta = \tan^{-1}(1/2)$ with the horizontal,with a speed of $36 \; km/h$. If the coefficient of friction of the surface is $1/\sqrt{3}$,then the power developed by the engine is:

Similar Questions

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:

$A$ solid hemisphere of weight $P$ rests with its curved surface in contact with a rough inclined plane. $A$ weight $Q$ is placed at some point on the rim of the hemisphere to keep its plane surface horizontal. Then its minimum coefficient of friction is

$A$ block of mass $2 \, kg$ slides down an inclined plane of inclination $30^{\circ}$. The coefficient of friction between the block and the plane is $0.5$. The contact force between the block and the plane is:

$A$ box is placed on an inclined plane and has to be pushed down. The angle of inclination is

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).

Vedclass Test Series

Exam Paper Generator

Online Exam Module