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UPSEE 2017 Solved Paper

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Question : 6 of 150

Marks: +1, -0

A time varying horizontal force (in Newton)

F=8\left|\sin(4\pi t)\right|

is acting on a stationary block of mass

2\,\text{kg}

as shown. Friction coefficient between the block and ground is

\mu=0.5

g=10\,\text{m/s}^2

. Then resulting motion of the block will be

It will oscillate
It remains stationary
It moves towards left
It moves towards right

Solution:

Given, Applied variable horizontal force

F=8\left|\sin(4\pi t)\right|\text{N}

Mass of block

(m)=2\,\text{kg}

Friction coefficient between block and ground

\mu=0.5

and

g=10\,\text{m/s}^2

We know that

maximum possible friction force,

F_{\text{friction max}}=\mu m g

=0.5\times 2\times 10

=10\,\text{N}

.......(i)

Maximum applied force

(F_{\text{applied max}})=8\times 1

=8\,\text{N}

........(ii)

[\because

maximum value of

\sin \theta

is 1

]

From equation (i) and (ii), we get

f_{\text{friction max}}>f_{\text{applied max}}

So, the block will remain stationary.

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