To solve the problem presented in the image, we need to calculate the rate at which the force does work on the block. This is also known as the power delivered by the force.

Given:

• Mass of the block: $m = 120 \, \text{kg}$
• Constant speed of the block: $v = 5.9 \, \text{m/s}$
• Force applied: $F = 140 \, \text{N}$
• Angle of the force above the horizontal: $\theta = 26^\circ$

Steps to calculate the power:

1. Work done by the force: The work done by the applied force is the component of the force in the direction of the motion (horizontal direction), which is $F \cos \theta$. Since the block is moving at constant speed, there is no net acceleration.

   Horizontal component of the force: $F_{\text{horizontal}} = F \cos \theta = 140 \, \text{N} \times \cos(26^\circ)$

2. Power formula: Power is the rate at which work is done. It is given by: $P = F_{\text{horizontal}} \times v$ where $v$ is the velocity.

Let's calculate it:

$F_{\text{horizontal}} = 140 \times \cos(26^\circ) \approx 140 \times 0.8988 = 125.832 \, \text{N}$

Now, calculate the power: $P = F_{\text{horizontal}} \times v = 125.832 \times 5.9 \approx 742.4 \, \text{W}$

Final Answer:

The rate at which the force does work on the block is approximately 742.4 watts.

Would you like more details on any step?

Here are 5 related questions to practice:

1. What would the power be if the angle were increased to 45°?
2. How does the speed of the block affect the power?
3. If the force applied was increased to 200 N, what would the new power be?
4. What is the significance of the angle in the context of work and power?
5. How would the power change if the block moved at a different constant speed?

Tip: Power is directly proportional to both force and velocity. Adjusting either affects how fast work is done.