Question

A shotput player throws a shotput of mass 3kg. If it crosses the top of wall of 2m height at a speed of 4m/s compute the total mehanical energy gained by shotput player. Take g=9.8m/s^2

Answer 1

machenical energy=ke+pe

K.E=1/2MV^2=1/2×3×4×4=24 J

P.E=m×g×h = 3×9.8×2= 58.8 J

M.E= K.E+P.E

= 24+58.8

= 72.8 JOULE

Answer 2

Given :

Mass (m) = 3 Kg

Velocity (v) = 4 m/sec

Height (h) = 2 m

To Find :

Total mechanical energy gained by shotput player.

Solution :

Potential energy of the shotput (PE) = mgh

                                                              = 3 × 9.8 × 2

                                                              = 58.8 J

Kinetic energy of the shotput  (KE) = $\frac{1}{2} mv^2$

                                                            = $\frac{1}{2}$ × 3 × (4)²

                                                            = 24 J

Total mechanical energy gained by shotput player = PE + KE

                                                                                       = 58.8 + 24

                                                                                       = 82.8 J

∴ The total mechanical energy gained by shotput player is 82.8 J