Carol and Bruno drag a box of mass 58.0 kg along a frictionless floor. Carol pushes the box with a force of 11.4 N at an angle of 40.0 downward from the horizontal. Bruno pulls the
box from the other side with a force of 11.0 N at an angle of 40.0 above the horizontal. What is the net work done on the box if the displacement of the box is 14.5 m?
w=fdcos
11.4(14.5)cos40 get that value and add that to 11.0(14.5)cos40
answer: 2.49 x 10^2 J
I didn't check calc work, but thinking is correct.
To calculate the net work done on the box, we first need to calculate the work done by Carol and Bruno separately and then add them together.
The formula to calculate work is W = F * d * cos(theta), where W is the work done, F is the force applied, d is the displacement, and theta is the angle between the force and displacement vectors.
Let's start with Carol's contribution:
Force applied by Carol (F_c) = 11.4 N
Displacement (d) = 14.5 m
Angle (theta_c) = 40.0 degrees (downward from the horizontal)
Using the formula, we can calculate the work done by Carol:
W_c = F_c * d * cos(theta_c)
= 11.4 N * 14.5 m * cos(40.0 degrees)
≈ 165.68 J
Next, let's calculate Bruno's contribution:
Force applied by Bruno (F_b) = 11.0 N
Displacement (d) = 14.5 m
Angle (theta_b) = 40.0 degrees (above the horizontal)
Using the formula, we can calculate the work done by Bruno:
W_b = F_b * d * cos(theta_b)
= 11.0 N * 14.5 m * cos(40.0 degrees)
≈ 157.19 J
Now, we can find the net work done by adding Carol's and Bruno's contributions:
Net work done (W_net) = W_c + W_b
≈ 165.68 J + 157.19 J
≈ 322.87 J
Therefore, the net work done on the box is approximately 322.87 J.