Question

A block of mass 6 kg is lying on a frictionless table. A force of 30 N is applied on
it for 12 seconds. Calculate its kinetic energy.​

Answer 1

Required Answer:

Given:

• Mass of the block (m) = 6kg

• Force applied ( F ) = 30N

• Time taken (t) = 12 seconds

To calculate:

• Kinetic energy ( K.E )

Calculation:

We know that,

• ${\underline {\boxed {\Large {\bf \gray { K.E = \dfrac{1}{2}m{v}^{2} } }}}}$

Here, velocity isn't given,so let us calculate the velocity first.

We know that,

• ${\underline {\boxed {\large {\sf \pink { F = ma } }}}}$

$\longmapsto$ 30 = 6 × a

$\longmapsto$ a = $\sf { \dfrac{30}{6} }$

$\longmapsto$ a = $\sf { 5m/{s}^{2}}$

Now,

By using 1st equation of:

• ${\underline {\boxed {\large {\sf \pink { v = u + at} }}}}$

Here, u (initial velocity) will be 0m/s.

$\longmapsto$ v = 0 + ( 5 × 12 )

$\longmapsto$ v = 0 + 60

$\longmapsto$ v = 60m/s

Therefore, velocity of the block is 60m/s.

By substituting all the values in K.E formula:

• ${\underline {\boxed {\large {\sf \pink { K.E = \dfrac{1}{2}m{v}^{2} } }}}}$

$\longmapsto$ K.E = $\sf{ \dfrac{1}{2} \times 6 \times {60}^{2} }$

$\longmapsto$ K.E = 1 × 3 × 3600

$\longmapsto$ K.E = 10,800 J

Therefore, kinetic energy of the block is 10,800 J.

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