If the contact surface between the 20-kg block and the ground is smooth, determine the power of force F when t = 4 s. Initially, the block is at rest

Respuesta :

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

[tex]F_{H}[/tex] = F cos0

[tex]F_{H}[/tex] = (30) x 4/5

[tex]F_{H}[/tex] = 24N

Now Calculate V using following formula

V = [tex]V_{0}[/tex] + at

[tex]V_{0}[/tex] = 0

a = [tex]F_{H}[/tex] / m

a = 24N / 20 kg

a = 1.2m / [tex]S^{2}[/tex]

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = [tex]F_{H}[/tex] x V

Power = 24 x 4.8

Power = 115.2 W