Question

An object of mass 5kg is resting on a smooth horizontal surface. A constant force of 20N is applied on the object horizontally for 10s. Calculate: (1) the work done by the force in 10s. (2) The K. E. of the object after 10s.

Answer 1

1)

Given,
mass = 5 kg
Horizontal net force = 20 N

We know that
F = ma

=> 20 = 5a

=> a = 20/5

=> a = 4 m/s²

Now acceleration = 4 m/s²
u = 0 m/s (since the body was at rest initially)
time = 10s

Now, from second equation of motion, we know that

$s = ut + \frac{1}{2} {at}^{2} \\ \\ = > s = 0 \times 10 + \frac{1}{2} \times 4 \times {10}^{2} \\ \\ = > s = 0 + 2 \times 100 \\ \\ = > s = 200$

Displacement = 200 m

Now work done = Force × Displacement

=> work done = 20 × 200

=> work done = 4000 Joules


2)

K.E (Kinetic energy) =
$\frac{1}{2} {mv}^{2}$


Now from first equation if motion we know that,

$v = u + at \\ \\ = > v = 0 + 4 \times 10 \\ \\ = > v = 40$
so v = 40 m/s

So K.E. =
$\frac{1}{2} {mv}^{2} \\ \\ = > \frac{1}{2} \times 5 \times {40}^{2} \\ \\ = > \frac{1}{2} \times 5 \times 1600 \\ \\ = > 5 \times 800 \\ \\ = > 4000$
Kinetic energy of the object = 4000 J

Hope it helps dear friend ☺️