Laws of Motion

A person is standing in an elevator. In which situation, he experiences weight loss?

An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10 ms^$$-$$2].

A monkey of mass $$50 \mathrm{~kg}$$ climbs on a rope which can withstand the tension (T) of $$350 \mathrm{~N}$$. If monkey initially climbs down with an acceleration of $$4 \mathrm{~m} / \mathrm{s}^{2}$$ and then climbs up with an acceleration of $$5 \mathrm{~m} / \mathrm{s}^{2}$$. Choose the correct option $$\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$$.

A force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be:

Given below are two statements : Statement I : An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable. Statement II : Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed. In the light of the above statements, choose the correct answer from the options given below :

The position vector of a particle related to time $t$ is given by $\vec{r}=\left(10 t \hat{i}+15 t^{2} \hat{j}+7 \hat{k}\right) m$ The direction of net force experienced by the particle is :

A body of mass $$500 \mathrm{~g}$$ moves along $$\mathrm{x}$$-axis such that it's velocity varies with displacement $$\mathrm{x}$$ according to the relation $$v=10 \sqrt{x} \mathrm{~m} / \mathrm{s}$$ the force acting on the body is:-

At any instant the velocity of a particle of mass $$500 \mathrm{~g}$$ is $$\left(2 t \hat{i}+3 t^{2} \hat{j}\right) \mathrm{ms}^{-1}$$. If the force acting on the particle at $$t=1 \mathrm{~s}$$ is $$(\hat{i}+x \hat{j}) \mathrm{N}$$. Then the value of $$x$$ will be:

A body of mass $4 \mathrm{~kg}$ experiences two forces $\vec{F}_1=5 \hat{i}+8 \hat{j}+7 \hat{k}$ and $\overrightarrow{\mathrm{F}}_2=3 \hat{i}-4 \hat{j}-3 \hat{k}$. The acceleration acting on the body is :

A cricket player catches a ball of mass $120 \mathrm{~g}$ moving with $25 \mathrm{~m} / \mathrm{s}$ speed. If the catching process is completed in $0.1 \mathrm{~s}$ then the magnitude of force exerted by the ball on the hand of player will be (in SI unit) :