Maricruz slides a 66 kg crate, m, up an inclined ramp 2.0 m long and attached to a platform 1.0 m above floor level, as shown in Figure 10-19. A 415 N force, F, parallel to the ramp, is needed to slide the crate up the ramp at a constant speed.

(a) How much work does Maricruz do in sliding the crate up the ramp?

(b) How much work would be done if Maricruz simply lifted the crate straight up from the floor to the platform?

(a) Well, it seems Maricruz is putting in some serious effort to slide that crate up the ramp. To calculate the work she does, we can use the formula:

Work = Force * Distance * cos(angle)

The angle here is the angle of the ramp, but since it's not given, let's assume it's perfectly inclined, meaning the angle is 0 degrees. That makes the cosine of 0 degrees equal to 1. So the formula simplifies to:

Work = Force * Distance

Plugging in the values, we get:

Work = 415 N * 2.0 m
= 830 N·m

(b) Now, if Maricruz decided to lift the crate straight up from the floor to the platform, she would have to overcome the gravitational force acting on it. The work done in this case would be given by:

Work = Force * Distance * cos(angle)

Since the force is the weight of the crate, we have:

Force = mass * gravity

So:

Work = (mass * gravity) * Distance * cos(angle)

Again, assuming the angle is 0 degrees, cosine of 0 is still 1, so the formula simplifies to:

Work = mass * gravity * Distance

Plugging in the values:

Work = 66 kg * 9.8 m/s^2 * 1.0 m
= 646.8 N·m

So, the work done by Maricruz to lift the crate straight up would be approximately 646.8 N·m. But hey, at least she gets a good workout!

To calculate the work done by Maricruz, we can use the formula:

Work = Force x Distance x cos(theta)

Where:
- Force is the parallel force exerted (415 N in this case)
- Distance is the length of the inclined ramp (2.0 m)
- theta is the angle between the force vector and the direction of motion (in this case, it would be the angle of the inclined ramp)

(a) To calculate the work done when sliding the crate up the ramp:

Since the crate is being slide up the inclined ramp, the angle between the force and the direction of motion is equal to the angle of the ramp, which can be determined using trigonometry.

The formula to calculate the angle, theta, is:

sin(theta) = opposite/hypotenuse

Where:
- the opposite side is the height of the ramp (1.0 m)
- the hypotenuse is the length of the ramp (2.0 m)

sin(theta) = 1.0/2.0
theta = arcsin(0.5)

Using a calculator, the value of theta is approximately 30.96 degrees.

Now, we can calculate the work done:

Work = Force x Distance x cos(theta)
= 415 N x 2.0 m x cos(30.96 degrees)

Calculating this, the work done by Maricruz to slide the crate up the ramp is approximately 716.73 Joules.

(b) To calculate the work done if Maricruz simply lifted the crate straight up from the floor to the platform:

When lifting the crate straight up, the angle between the force and the direction of motion is 90 degrees, which means cos(90 degrees) = 0.

Therefore, the work done when lifting the crate straight up would be zero, as no work is done in the direction of the force applied.

So the work done would be zero.

To solve this problem, we need to use the concept of work. Work is defined as the product of the force applied on an object and the displacement of the object in the direction of the force.

(a) To calculate the work done by Maricruz in sliding the crate up the ramp, we can use the formula:

Work = Force * Distance * cosine(theta)

Given:
Mass of crate (m) = 66 kg
Length of ramp (distance) = 2.0 m
Height of platform (h) = 1.0 m
Force required (F) = 415 N

To find the angle theta, we can use trigonometry. The ramp and the horizontal plane form a right triangle, so we can use the equation:

sin(theta) = h / d

where h is the height of the platform and d is the length of the ramp. Plugging in the values, we have:

sin(theta) = 1.0 / 2.0 = 0.5

Now we can find theta by using the inverse sine function:

theta = sin^(-1)(0.5)

Using a calculator, we find:

theta ≈ 30°

Plugging the values into the work formula, we get:

Work = 415 N * 2.0 m * cos(30°)

To find cos(30°), we can use a calculator again:

cos(30°) ≈ 0.866

Now we can calculate the work:

Work = 415 N * 2.0 m * 0.866

Work ≈ 719 J

Therefore, Maricruz does approximately 719 Joules of work in sliding the crate up the ramp.

(b) If Maricruz were to lift the crate straight up from the floor to the platform, the work done would be given by the formula:

Work = Force * Distance

The force required would be equal to the weight of the crate, which is given by:

Weight = mass * acceleration due to gravity

Weight = 66 kg * 9.8 m/s^2 (approximately)

Weight ≈ 646.8 N

Now we can calculate the work done:

Work = 646.8 N * 1.0 m

Work ≈ 646.8 J

Therefore, if Maricruz were to lift the crate straight up from the floor to the platform, she would need to do approximately 646.8 Joules of work.

Fp = 415N. = Force parallel to the ramp

a. W = Fp*d = 415 * 2 = 830 Joules.

b. Wc = mg = 66kg * 9.8N/ikg = 647N. =
Weight of crate.

Work = Fc*h = 647 * 1 = 647 Joules.