A traveler pulls on a suitcase strap at an angle 36 above the horizontal. If 555 J of work are done by the strap while moving the suitcase a horizontal distance of 15 m, what is the tension in the strap?
See previous post: Mon, 10-27-14, 2:13 PM.
To find the tension in the strap, we can use the work-energy principle. The work done by the strap is equal to the change in the suitcase's kinetic energy.
First, let's calculate the change in kinetic energy by using the work-energy principle:
Work done = Change in kinetic energy
Given that the work done is 555 J.
555 J = ΔK
Now, let's find the change in kinetic energy by using the formula for kinetic energy:
Kinetic energy = (1/2)mv^2
Since the suitcase is moving horizontally, its velocity in the vertical direction is zero. Therefore, the only component of velocity that affects its kinetic energy is the horizontal component.
Let's assume the tension in the strap is T, and the mass of the suitcase is m.
Since the angle between the force applied (tension) and the direction of motion is 36 degrees, we can find the horizontal component of the force by multiplying the tension by the cosine of the angle:
Force in the horizontal direction = T * cos(36)
Now, using Newton's second law, we can relate the horizontal force, mass, and acceleration:
Force in the horizontal direction = m * a
The acceleration in the horizontal direction is given by:
a = v^2 / d
where v is the horizontal component of the velocity and d is the horizontal distance traveled.
Given that the horizontal distance traveled is 15 m, and the vertical velocity component is zero, the horizontal component of the velocity is equal to the total velocity.
To find the total velocity, we can use the Pythagorean theorem:
v^2 = (vertical velocity component)^2 + (horizontal velocity component)^2
Since the vertical velocity component is zero, the total velocity simplifies to the horizontal velocity component.
Now we have all the information we need to solve for the tension in the strap:
555 J = (1/2)m * (horizontal velocity component)^2
horizontal velocity component = v
T * cos(36) = m * v^2 / 15^2
T * cos(36) = m * v^2 / 225
Since we are trying to find the tension in the strap, we can isolate T:
T = (m * v^2 / 225) / cos(36)
Finally, substitute the given values and calculate the tension in the strap.