Question

Question 14
A 2 kg block is kept at rest on a rough horizontal surface having coefficient of static and kinetic friction as 0.2
and
0.1
respectively. A force F = 2t starts acting on the object (where it is in seconds). Find 2v if v is speed of
the block at t = 3 sec.
​

Answer 1

2v = 3 m/s

Where v is the speed of the object after 3 seconds

Explanation:

Given:

Mass of the block m = 2 kg

Coefficient of static friction $\mu_s=0.2$

Coefficient of kinetic friction $\mu_k=0.1$

Force acting on the block F = 2t

Where t is in seconds

To find out:

Twice the speed of the block after 3 seconds

Solution:

Initially when the block is at rest then the static friction force will act on the block

Thus, in limiting condition

$F=\mu_sN$

$\implies F=\mu_smg$

$\implies 2t=0.2\times 2\times 10$

$\implies t=2$ seconds

Therefore, the block will start moving after 2 seconds

We have to find the speed of the block after 1 second

If the velocity of the block after t second is u and its acceleration is a

Then

$F-\mu_kN=ma$

$\implies 2t-0.1\times 2\times 10=2a$

$\implies a=t-1$

$\implies \frac{du}{dt}=t-1$

$\implies du=(t-1)dt$

$\implies \int_0^u=\int_2^3(t-1)dt$

$\implies u\Bigr|_0^u=(\frac{t^2}{2}-t)\Bigr|_2^3$

$\implies u=(\frac{3^2}{2}-3)-(\frac{2^2}{2}-2)$

$\implies u=\frac{3}{2}$ m/s

Therefore, speed after 3 seconds

$v=|u|$

$\implies v=\frac{3}{2}$ m/s

Twice the speed

$2v=2\times \frac{3}{2}$

$\implies v=3$ m/s

Hope this answer is helpful.

Know More:

Q: a block of mass 2kg is on a horizontal surface .the coefficient of static and kinetic frictions are 0.6 n0.2 the minimum horizontal force required to start the motion is applied and if it is continued the velocity acquired by the body at the end of 2nd second if g=10 is

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Click Here: https://brainly.in/question/44366

Answer 2

Given:

A 2 kg block is kept at rest on a rough horizontal surface having coefficient of static and kinetic friction as 0.2 and 0.1 respectively. A force F = 2t starts acting on the object (where it is in seconds).

To find:

Value of 2v , when v is velocity at t = 3 sec.

Calculation:

First of all, we need to find out the time at which the block just starts moving , also called CRITICAL CONDITION.

$\sf \therefore \: F = \mu_{s} \times (mg)$

$\sf \implies \: 2t = 0.2 \times (2 \times 10)$

$\sf \implies \: t = 2 \: sec$

Now , after 2 secs , the net force experienced by the object will be :

$\sf \therefore \: F_{net} = F - \mu_{k} \times (mg)$

$\sf \implies \: F_{net} = 2t - (0.1\times 2 \times 10)$

$\sf \implies \: F_{net} = 2t - 2$

$\sf \implies \: a_{net} = \dfrac{2t - 2}{m}$

$\sf \implies \: a_{net} = \dfrac{2t - 2}{2}$

$\sf \implies \: a_{net} = t - 1$

$\sf \implies \: \dfrac{dv}{dt} = t - 1$

$\displaystyle \sf \implies \: \int dv = \int(t - 1) \: dt$

Putting limits :

$\displaystyle \sf \implies \: \int_{0}^{v} dv = \int_{2}^{3}(t - 1) \: dt$

$\displaystyle \sf \implies \: v - 0 = \bigg \{\dfrac{{t}^{2}}{2} - t \bigg \}_{2}^{3}$

$\displaystyle \sf \implies \: v = ( \dfrac{{3}^{2} - {2}^{2}}{2} ) - (3 - 2)$

$\displaystyle \sf \implies \: v = \dfrac{5}{2}- 1$

$\displaystyle \sf \implies \: v = \dfrac{3}{2}$

$\displaystyle \sf \implies \: 2v = 3$

So, value of 2v will be 3 m/s.