Question

6. A block of mass 2 kg is placed on a truck as shown in figure. The coefficient of kinetic friction between
the block and surface is 0.5. The truck starts from rest and moves with acceleration 8 m/s2. After how much time the block fall off the truck

1) 4s
2) 3s
3) 2s
4) 1.5 s​

Answer 1

Free Body Diagram of the block is given below.

$\setlength{\unitlength}{0.9mm}\begin{picture}(5,5)\thicklines\put(0,0){\framebox(10,10){\sf{2\ kg}}}\put(0,5){\vector(-1,0){20}}\put(-26,4){\sf{2a}}\put(10,-0.1){\vector(1,0){10}}\put(21,-1){\sf{f}}\put(5,0){\vector(0,-1){15}}\put(1,-19){\sf{20\ N}}\put(5,10){\vector(0,1){15}}\put(4,26){\sf{R}}\end{picture}$

Take $\sf{g=10\ m\,s^{-2}.}$

Since net vertical force is zero, the reaction acting on the block is,

$\sf{\longrightarrow R=20\ N}$

As the truck accelerates forward the block experiences a pseudo force in opposite direction as that of motion of truck.

Thus frictional force is acting on the block opposite to its motion, i.e., along the direction of motion of truck.

The magnitude of frictional force,

$\sf{\longrightarrow f=\mu R}$

$\sf{\longrightarrow f=0.5\times20}$

$\sf{\longrightarrow f=10\ N}$

Net horizontal force acting on the block (leftwards),

$\sf{\longrightarrow 2a_{net}=2a-f}$

$\sf{\longrightarrow 2a_{net}=2\times8-10\ }$

$\sf{\longrightarrow 2a_{net}=6\ N}$

And net horizontal acceleration,

$\sf{\longrightarrow a_{net}=\dfrac{6}{2}}$

$\sf{\longrightarrow a_{net}=3\ m\,s^{-2}}$

The block has to move 6 metres leftwards (along the direction of net acceleration) to fall off the truck.

The time taken to fall off the truck is given by second equation of motion as,

$\sf{\longrightarrow s=ut+\dfrac{1}{2}\,a_{net}t^2}$

Since the block starts from rest,

$\sf{\longrightarrow 6=(0)t+\dfrac{1}{2}\,(3)t^2}$

$\sf{\longrightarrow 6=\dfrac{3}{2}\,t^2}$

$\sf{\longrightarrow\underline{\underline{t=2\ s}}}$

Hence (3) is the answer.

Answer 2

Answer:

7

Take \sf{g=10\ m\,s^{-2}.}g=10 ms

−2

.

Since net vertical force is zero, the reaction acting on the block is,

\sf{\longrightarrow R=20\ N}⟶R=20 N

As the truck accelerates forward the block experiences a pseudo force in opposite direction as that of motion of truck.

Thus frictional force is acting on the block opposite to its motion, i.e., along the direction of motion of truck.

The magnitude of frictional force,

\sf{\longrightarrow f=\mu R}⟶f=μR

\sf{\longrightarrow f=0.5\times20}⟶f=0.5×20

\sf{\longrightarrow f=10\ N}⟶f=10 N

Net horizontal force acting on the block (leftwards),

\sf{\longrightarrow 2a_{net}=2a-f}⟶2a

net

=2a−f

\sf{\longrightarrow 2a_{net}=2\times8-10\ }⟶2a

net

=2×8−10

\sf{\longrightarrow 2a_{net}=6\ N}⟶2a

net

=6 N

And net horizontal acceleration,

\sf{\longrightarrow a_{net}=\dfrac{6}{2}}⟶a

net

=

2

6

\sf{\longrightarrow a_{net}=3\ m\,s^{-2}}⟶a

net

=3 ms

−2

The block has to move 6 metres leftwards (along the direction of net acceleration) to fall off the truck.

The time taken to fall off the truck is given by second equation of motion as,

\sf{\longrightarrow s=ut+\dfrac{1}{2}\,a_{net}t^2}⟶s=ut+

2

1

a

net

t

2

Since the block starts from rest,

\sf{\longrightarrow 6=(0)t+\dfrac{1}{2}\,(3)t^2}⟶6=(0)t+

2

1

(3)t

2

\sf{\longrightarrow 6=\dfrac{3}{2}\,t^2}⟶6=

2

3

t

2

\sf{\longrightarrow\underline{\underline{t=2\ s}}}⟶

t=2 s

Hence (3) is the answer.