A bullet of mass $$m$$ fired at $$30^{\circ}$$ to the horizontal leaves the barrel of the gun with a velocity $$v$$. The bullet hits a soft target at a height $$h$$ above the ground while it is moving downward and emerge out with half the kinetic energy it had before hitting the target.
Which of the following statements are correct in respect of bullet after it emerges out of the target?
Two blocks $$M_1$$ and $$M_2$$ having equal mass are free to move on a horizontal frictionless surface. $$M_2$$ is attached to a massless spring as shown in figure. Initially $$M_2$$ is at rest and $$M_1$$ is moving toward $$M_2$$ with speed $$v$$ and collides head-on with $$M_2$$.
A rough inclined plane is placed on a cart moving with a constant velocity $$u$$ on horizontal ground. A block of mass $$M$$ rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?
Consider the adjacent diagram. As the block M is at rest.
Hence, f = frictional force = $$Mg\sin\theta$$
The force of friction acting between the block and incline opposes the tendency of sliding of the block. Since, block is not in motion, therefore, no work is done by the force of friction. Hence, no dissipation of energy takes place.
Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?
When the elevator is descending, then electric power is required to prevent it from falling freely under gravity.
Also, as the weight inside the elevator increases, its speed of descending increases, therefore, there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with large velocity.
A body is being raised to a height $h$ from the surface of earth. What is the sign of work done by
(a) applied force and
(b) gravitational force?
(a) Force is applied on the body to lift it in upward direction and displacement of the body is also in upward direction, therefore, angle between the applied force and displacement is $$\theta=0^{\circ}$$
$$\therefore$$ Work done by the applied force
$$W=F s \cos \theta=F s \cos 0^{\circ}=F s \left(\because \cos 0^{\circ}=1\right)$$
i.e., $$\qquad W=$$ Positive
(b) The gravitational force acts in downward direction and displacement in upward direction, therefore, angle between them is $$\theta=180^{\circ}$$.
$$\therefore$$ Work done by the gravitational force
$$W=F s \cos 180^{\circ}=-F s \quad\left(\because \cos 180^{\circ}=1\right)$$