# How to do this problem??

Not able to correctly understand the graph. Is the relation F=40t?????

Note by Rajath Krishna R
5 years, 10 months ago

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ohhhh very small pic

- 5 years, 10 months ago

This is the question........ A 20 kg block is originally at rest on a horizontal surface for which the coefficient of friction is 0.6. If a horizontal force is applied such that it varies with time as shown. Determine the speed of block in 10 sec.

- 5 years, 10 months ago

is the answer 58.8 m/s ?

- 5 years, 10 months ago

noop.....

- 5 years, 10 months ago

6 m/s ?????

- 5 years, 10 months ago

Is the answer 49 m/s ? (assuming g=10 m/s^2)

- 5 years, 10 months ago

what's the need of g here????

- 5 years, 10 months ago

You need the value of g to know when will the body start moving (3 seconds in this case) Because for the body to move, F>= 0.6Mg i.e. F>=120. Hence the body starts moving after 3 secs.

- 5 years, 10 months ago

yeah i forgot about it.............I too did the same way...

- 5 years, 10 months ago

Oh!! i got the question wrong!! the answer must be 24 m\s. (with g=10 m\s^2)

- 5 years, 10 months ago

yeah its right......how did u do it..................

- 5 years, 10 months ago

@Rajath there are two ways of doing this. Actually both are different approaches having same meaning. If you are familiar with basic integration you can try integrating the net force wrt time to get the change in momentum hence you can find the velocity. Or you can first try to sketch the graph of net force versus time and find it's area b/w t=3 to t=10.(Both are same methods actually as integration is nothing but area bounded). The area gives you the change in momentum therefore you can calculate velocity. But remember that we are not finding area under the graph in the question. We have to sketch our own graph of net force (Just the given graph pull it downwards by 120 )

if you do that properly the area comes out to be 480. hence the velocity would be 480/20= 24 m/s.

- 5 years, 10 months ago