Question

A body of mass 20kg at rest. A force of 5n is applied on it. Calculate the work done in the first second

Answer 1

Answer:0.625J

Explanation:m=20 kg

F=5N

T=1s.

u=0

=S=ut+1/2at^2

=0×1+1/2at^2 = 1/2at^2

F=m.a = a=F/m= 1/2×F/m×t^2

=W=F.S

=F×1/2×F/m×t^2

=5×1/2×5/20×1^2

=5/8

= 0.625J

Answer 2

The work done by a force is defined to be the product of a component of the force in the direction of the displacement and the magnitude of this displacement. Formula. Work can be calculated by multiplying Force and Distance in the direction of force as follows. $W = F \times d$. Unit.

$m=20kg\\F=5N$

$t=1 sec$

$A=\frac{F}{m}=\frac{5}{20}=\frac{1}{4} m/s^2$

$t=1sec$

Now, work done $=FS$

$s=ut+\frac{1}{2} ut^2$

$u=1.125m$

$F=5\times 1.125=5.625J$

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