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:

115.2 W

Explanation:

The computation is shown below:

As we know that

Power = F . v

$F_H = F cos \theta$

$F_H = 30 \frac{4}{5}$

$F_H = 24N$

Now we solve for V

$V = V_0 + at$            a = 24N ÷ 20Kg

But V_0 = 0          a = 1.2 m/s^2

F_H = ma             V = 0 + (1.2) (4)

a = F_H ÷ m        V = 4.8 m/s

Therefore

Power = F_Hv

= (24) (4.8)

= 115.2 W

By applying the above formuals we can get the power