Question

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

Answer 1

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

$F_{H}$ = F cos0

$F_{H}$ = (30) x 4/5

$F_{H}$ = 24N

Now Calculate V using following formula

V = $V_{0}$ + at

$V_{0}$ = 0

a = $F_{H}$ / m

a = 24N / 20 kg

a = 1.2m / $S^{2}$

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = $F_{H}$ x V

Power = 24 x 4.8

Power = 115.2 W