A cable exerts a constant upward tension of magnitude 1.82 ✕ 104 N on a 1.60 ✕ 103 kg elevator as it rises through a vertical distance of 2.90 m. Find the work done by the tension force on the elevator (in J). Find the work done by the force of gravity on the elevator (in J).
*Hint: Apply the equation for the work done by a constant force during a linear displacement, W = (F cos(θ)) d.
work = force * distance moved in direction of force
for tension:
W = 1.82 * 10^4 N * 2.90 m =
for gravity (force down, motion up so -)
W = -1.60*10^3 * 9.81 * 2.90 =
1. 52780 J
2. -15696
Well, let's calculate the work done by the tension force on the elevator first.
We have the magnitude of the tension force as 1.82 ✕ 104 N, and the vertical distance the elevator rises as 2.90 m. Since the cable exerts an upward tension, the angle θ between the force and the displacement is 0 degrees (or cos(0) = 1).
Using the equation for work, W = (F cos(θ)) d, we can substitute in the values:
W = (1.82 ✕ 104 N)(cos(0))(2.90 m)
Since cos(0) = 1, we can simplify further:
W = (1.82 ✕ 104 N)(1)(2.90 m)
W = 5.28 ✕ 104 J
So, the work done by the tension force on the elevator is 5.28 ✕ 104 J.
Now let's find the work done by the force of gravity on the elevator.
The force of gravity acting on the elevator is given by the weight, which is the mass (1.60 ✕ 103 kg) multiplied by the acceleration due to gravity (9.8 m/s²).
So, the force of gravity Fg = (1.60 ✕ 103 kg)(9.8 m/s²)
Now, we need to find the angle θ between the force of gravity and the displacement. Since both the force and the displacement are vertical, the angle between them is 180 degrees (or cos(180) = -1).
Using the equation for work, W = (F cos(θ)) d, we can substitute in the values:
W = (Fg)(cos(180))(2.90 m)
Since cos(180) = -1, we can simplify further:
W = (Fg)(-1)(2.90 m)
W = -4.67 ✕ 104 J
So, the work done by the force of gravity on the elevator is -4.67 ✕ 104 J.
Hey, at least gravity is keeping us grounded!
To find the work done by the tension force on the elevator, we'll use the equation for the work done by a constant force during a linear displacement:
W = (F cos(θ)) d
Here, F is the magnitude of the tension force, θ is the angle between the force and the displacement, and d is the displacement.
In this case, the tension force is acting upward, so the angle between the force and the displacement is 0 degrees (cos(0) = 1).
Given:
Tension force (F) = 1.82 ✕ 104 N
Displacement (d) = 2.90 m
Plugging in the values, we get:
W = (1.82 ✕ 104 N) × (cos(0)) × (2.90 m)
W = (1.82 ✕ 104 N) × (1) × (2.90 m)
W = 1.82 ✕ 104 N × 2.90 m
W = 5.278 ✕ 104 J
So, the work done by the tension force on the elevator is 5.278 ✕ 104 J.
Now, let's find the work done by the force of gravity on the elevator.
Since the elevator is being lifted in a vertical direction, the force of gravity is acting in the opposite direction to the displacement. So, the angle between the force of gravity and the displacement is 180 degrees (cos(180) = -1).
The weight of the elevator can be found using the formula:
Weight = mass × gravity
Given:
Mass of the elevator = 1.60 ✕ 103 kg
Acceleration due to gravity = 9.8 m/s^2
Weight = (1.60 ✕ 103 kg) × (9.8 m/s^2)
Weight = 1.568 ✕ 104 N
Plugging in the values, we get:
W = (1.568 ✕ 104 N) × (cos(180)) × (2.90 m)
W = (1.568 ✕ 104 N) × (-1) × (2.90 m)
W = -1.568 ✕ 104 N × 2.90 m
W = -4.5432 ✕ 104 J (note that the negative sign indicates the direction of the work done by the force of gravity)
So, the work done by the force of gravity on the elevator is -4.5432 ✕ 104 J.
To find the work done by the tension force on the elevator, we can use the equation for the work done by a constant force during a linear displacement, which is given as:
W = (F cos(θ)) d
Where:
W is the work done,
F is the magnitude of the force,
θ is the angle between the force and the direction of displacement,
and d is the displacement.
In this case, the force exerted by the cable is upwards, so the angle between the force and the direction of displacement is 0 degrees (or cos(0) = 1). Therefore, we can simplify the equation to:
W = F d
Plugging in the given values:
F = 1.82 ✕ 10^4 N
d = 2.90 m
W = (1.82 ✕ 10^4 N)(2.90 m)
W ≈ 5.28 ✕ 10^4 J
Therefore, the work done by the tension force on the elevator is approximately 5.28 ✕ 10^4 J.
To find the work done by the force of gravity on the elevator, we can use the same equation and consider that the force of gravity always acts in the downward direction. In this case, the angle between the force of gravity and the direction of displacement is again 0 degrees (or cos(0) = 1). Therefore, using the same equation as above:
W = F d
The force of gravity can be calculated using the formula F = m g, where m is the mass of the elevator and g is the acceleration due to gravity. Plugging in the given values:
m = 1.60 ✕ 10^3 kg
g = 9.8 m/s^2
F = (1.60 ✕ 10^3 kg)(9.8 m/s^2)
F = 1.57 ✕ 10^4 N
Now we can calculate the work done by the force of gravity on the elevator:
W = (1.57 ✕ 10^4 N)(2.90 m)
W ≈ 4.55 ✕ 10^4 J
Therefore, the work done by the force of gravity on the elevator is approximately 4.55 ✕ 10^4 J.