A block of mass m = 364 g is dragged with a string across a rough horizontal table. The string tension is T = 3.59 N, and it pulls upward at an angle of ¦Õ = 47.0¡ã with the horizontal. At one particular instant the block is moving at a speed of v = 6.50 m/s. The coefficient of kinetic friction between the block and the table is ¦Ìk = 0.639.
What is the power supplied to the block by the string tension?
What is the power supplied by the force of friction?
To find the power supplied to the block by the string tension, we can use the formula for power:
Power = Force x Velocity
First, let's find the horizontal component of the tension force. We can use the trigonometric relationship:
cos(θ) = adj / hypotenuse
cos(47°) = horizontal component of tension force / tension force
Solving for the horizontal component of tension force:
horizontal component of tension force = T * cos(47°)
Now, we can find the power supplied to the block by the string tension:
Power = (T * cos(47°)) * v
Substituting the given values:
Power = (3.59 N * cos(47°)) * 6.50 m/s
To solve this equation, we need to evaluate the cosine of 47° and then multiply it by 3.59 N and 6.50 m/s. Evaluating the cosine of 47° gives us approximately 0.682.
Power ≈ (3.59 N * 0.682) * 6.50 m/s
Power ≈ 15.36 Watts
Therefore, the power supplied to the block by the string tension is approximately 15.36 Watts.
To find the power supplied by the force of friction, we can use the formula:
Power = Force x Velocity
The force of friction can be calculated using the equation:
Force of friction = coefficient of kinetic friction * normal force
The normal force can be calculated using the equation:
Normal force = mass * gravitational acceleration
In this case, mass = 364 g = 0.364 kg and gravitational acceleration is approximately 9.8 m/s².
Now, we can find the force of friction:
Force of friction = 0.639 * (0.364 kg * 9.8 m/s²)
Substituting and solving for the force of friction:
Force of friction = 0.639 * (0.364 kg * 9.8 m/s²)
Force of friction ≈ 2.25 N
Finally, we can find the power supplied by the force of friction:
Power = (2.25 N) * 6.50 m/s
Power ≈ 14.63 Watts
Therefore, the power supplied by the force of friction is approximately 14.63 Watts.