An $80\, kg$ person is parachuting and is experiencing a downward acceleration of $2.8\, m/s^2$. The mass of the parachute is $5\, kg$. The upward force on the open parachute is ........... $N$ (Take $g = 9.8\, m/s^2$)

The arrangement of a two-block system is as shown. Assuming the surfaces to be frictionless,the net force acting on the $1 \, kg$ and $2 \, kg$ blocks are,respectively:

Two blocks are in contact on a frictionless table. One has mass $m$ and the other $2\,m$. $A$ force $F$ is applied on $2\,m$ as shown in the figure. Now,the same force $F$ is applied from the right on $m$. In the two cases respectively,what is the ratio of the contact force between the two blocks?

$A$ force $F = 10t$ acts on a block of mass $m = 10 \, kg$ as shown in the figure. The coefficient of friction is $\mu = 0.4$. Find the time $t$ when the block loses contact with the surface.

$A$ block of mass $m$ is kept on a platform which starts from rest with constant acceleration $g/2$ upward,as shown in the figure. Work done by the normal reaction on the block in time $t$ is

$A$ block of mass $m$ slides on a wooden wedge,which in turn slides backward on a horizontal surface. The acceleration of the block with respect to the wedge is: Given $m = 8 \, kg, M = 16 \, kg$. Assume all surfaces shown in the figure to be frictionless.