A block of mass 'm' with an initial kinetic e | vedclass

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

Question

(D) Let the body slide up to a distance '$s$' before coming to rest.The height reached is $h = s \sin 	heta$.According to the work-energy theorem,the

Vedclass · Accepted Answer

$\frac{\mu E \cos 	heta}{\sin 	heta+\mu \cos 	heta}$

Vedclass · Answer

(D) Let the body slide up to a distance '$s$' before coming to rest.The height reached is $h = s \sin 	heta$.According to the work-energy theorem,the total work done by all forces equals the change in kinetic energy:$W_{friction} + W_{gravity} = K_f - K_i$$-\mu mg \cos 	heta \cdot s - mg \sin 	heta \cdot s = 0 - E$$\mu mgs \cos 	heta + mgs \sin 	heta = E$$s(mg(\mu \cos 	heta + \sin 	heta)) = E$$s = \frac{E}{mg(\mu \cos 	heta + \sin 	heta)}$The work done against friction is $W_{against\ friction} = |W_{friction}| = \mu mg \cos 	heta \cdot s$.Substituting the value of '$s$':$W_{against\ friction} = \mu mg \cos 	heta \cdot \left( \frac{E}{mg(\mu \cos 	heta + \sin 	heta)} ight)$$W_{against\ friction} = \frac{\mu E \cos 	heta}{\sin 	heta + \mu \cos 	heta}$

$A$ block of mass '$m$' with an initial kinetic energy '$E$' moves up an inclined plane of inclination '$\theta$'. If '$\mu$' is the coefficient of friction between the plane and the body,the work done against friction before coming to rest is

Similar Questions

$A$ ball is falling freely from a height. When it reaches $10 ~m$ height from the ground,its velocity is $v_0$. It collides with the ground,loses $50 \%$ of its energy,and rises back to a height of $10 ~m$. Then the velocity $v_0$ is (in $~m / s$)

$A$ child swings on a swing such that the minimum and maximum heights from the ground are $0.75 \ m$ and $2 \ m$ respectively. The maximum velocity of the swing is ...... $ms^{-1}$.

Select the false statement.

Two balls at the same temperature collide. What is conserved?

$300\, J$ of work is done in sliding a $2\, kg$ block up an inclined plane of height $10\, m.$ Work done against friction is.....$J$ (Take $g = 10\, m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module