A body of mass 2 kg having an initial speed of 11 m/s accelerated uniformly at 2 m/s for 5 seconds. What will be the final kinetic energy?

Open in App

Solution

From the first equation of motion:
$v = u + a t v = 11 + 2 \times 5 v = 11 + 10 = 21 m / s$
Kinetic Energy $= \frac{1}{2} m v^{2} = \frac{1}{2} \times 2 \times 21 \times 21 = 441 J$

Suggest Corrections

Similar questions

Q. A block of mass

2.00 k g

moving at a speed of

10.0 m / s

accelerates at

3.00 m / s^{2}

for

5.00 s

. Compute its final kinetic energy.

Q. A body accelerates uniformly from rest to a velocity of

1 m s^{- 1}

in 15 second. In what time the kinetic energy of the body will be

\frac{2}{9} J

.[Take mass of body as 1kg]:

A body of mass 2 kg is thrown upward with an energy 490 J. The height at which its kinetic energy would become half of its initial kinetic energy will be $[g = 9.8 m / s^{2}]$

Q. A body of mass

2 k g

is thrown upward with initial velocity

20 m / s

. After

2 s

find its kinetic energy will be:

(g = 10 m / s^{2})

A car has initial velocity of 5 m/s is increasing its acceleration uniformly by 2 m/s2. At what distance will be the car after 12 seconds ?

Join BYJU'S Learning Program

A body of mass 2 kg having an initial speed of 11 m/s accelerated uniformly at 2 m/s for 5 seconds. What will be the final kinetic energy?

From the first equation of motion: v=u+atv=11+2×5v=11+10=21m/s Kinetic Energy =12mv2=12×2×21×21=441J

From the first equation of motion:
$v = u + a t v = 11 + 2 \times 5 v = 11 + 10 = 21 m / s$
Kinetic Energy $= \frac{1}{2} m v^{2} = \frac{1}{2} \times 2 \times 21 \times 21 = 441 J$