Two forces, of magnitudes F_1 = 80.0 N and F_2 = 25.0 N, act in opposite directions on a block, which sits atop a frictionless surface. Initially, the center of the block is at position x_i = -4.00 cm. At some later time, the block has moved to the right, and its center is at a new position, x_f = 2.00 cm. Find the work W_1 done on the block by the force of magnitude F_1 = 80.0 N as the block moves from x_i = -4.00 cm to x_f = 2.00 cm.
work doneby F1=80*.06 joules
work doneby F2=-25*.06
When F1 = 1500 lb upward and F2 = 800 lb downward, the magnitude of the resultant and its direction would be
To find the work done by the force F_1 as the block moves from x_i to x_f, we can use the formula:
W = F * d * cosθ
where W is the work done, F is the magnitude of the force, d is the displacement, and θ is the angle between the force and the displacement.
In this case, the force F_1 acts in the positive direction (to the right) and the displacement is also in the positive direction (from left to right). Therefore, cosθ = 1.
The magnitude of the force F_1 is given as 80.0 N. The displacement d is the difference between the final and initial positions, which is:
d = x_f - x_i = 2.00 cm - (-4.00 cm) = 6.00 cm
Now, we need to convert the displacement from centimeters to meters:
d = 6.00 cm = 6.00 cm * (1 m / 100 cm) = 0.06 m
Substituting the values into the formula, we have:
W_1 = F_1 * d * cosθ
= 80.0 N * 0.06 m * 1
= 4.80 Joules
Therefore, the work done on the block by the force F_1 is 4.80 Joules.
To find the work done by the force of magnitude F1 as the block moves from xi to xf, we can use the equation:
W = F · d · cosθ
where W is the work done, F is the magnitude of the force, d is the displacement, and θ is the angle between the force and the displacement.
In this case, the force F1 is acting in the negative x-direction and the displacement is from xi = -4.00 cm to xf = 2.00 cm, which means the displacement is in the positive x-direction.
First, let's calculate the displacement d:
d = xf - xi
= 2.00 cm - (-4.00 cm)
= 6.00 cm or 0.06 m (converting to meters)
Next, we need to find the angle θ between the force F1 and the displacement. Since the force and displacement are in opposite directions, the angle θ is 180 degrees or π radians.
Now we can calculate the work W1:
W1 = F1 · d · cosθ
= 80.0 N · 0.06 m · cos(180°)
To evaluate the cosine of 180 degrees, we need to convert it to radians:
cos(180°) = cos(π rad)
Using the value for π (pi) as 3.14:
W1 = 80.0 N · 0.06 m · cos(π rad)
= 80.0 N · 0.06 m · (-1)
= -2.40 J
Therefore, the work W1 done on the block by the force of magnitude F1 is -2.40 Joules.