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Question: A block of mass 2kg placed on a long frictionless horizontaltable is pulled horizontally by a constant force F. It is found to move 10m in the first two seconds. Find magnitude of F.

Posted by: Ashmi on 18.11.2020

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Answer:
S=ut+1/2at^2 10 =1/2*a*4 a=5 F=2*5=10N
Answer:

Given, s = 10 m, t = 2 sec, u = 0, m = 2 kg

Now, s = ut + 1/2.a.t^2

So, 10 = 1/2.a.4 = 2a

So, a = 10/2 = 5 m/(s^2)

Again, F = m.a

Hence, F = 2x5 N = 10 N    ... (Ans)

Answer:

We know that s=ut+1/2at2,here s=distance=10m,u=initial velocity=10/2=5m/s,a=acceleration,,t=time =2sec

10=o*2+1/2a2*

a=5m/s2

Force=m*a=2kg*5m/s2=10N

Answer:

The block moves a ditance of 10 metres in 2 seconds under the action of force.As there is no friction,we can assume that the block is moving with uniform  acceleration. So the distsnce trasvelled by the block with uniform acceleration is given by the formula s=ut+(1/2)*a*t^2, whwre s= distance ravelled by the block= 10  metres , u= ininitial velocity= 0         ( because block was under rest before application of force) ,  a= uniform acceleration of the body, t= time taken to attain the final velocity= 2 seconds  and m=mass of the block= 2Kgs. Applying these values in the equation,we get                  10 = (1/2)*a*2*2= 2a which gives a=5 metres/sq.sec.Therfore Force f= m*a= 2*5= 10  Newtons

Ans: 10 Newtons

Answer:

s=ut+1/2 at^2

=> a=5 m/s^2

now F=ma= 2x5=10N

Answer:

F=ma

Since F is constant, a is also constant.Therefore from Kinematics:

S=ut+21​at2

10=0+21​a×4 (Since object is at rest initially)

a=210​=5m/s2

F=2×5=10N

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