Question

A girl pulls her 15-kg wagon along a flat sidewalk by applying a 10-N force at 37° to the horizontal. Assume that friction is negligible and that the wagon starts from rest. (a) How much work does the girl do on the wagon in the first 2.0 s. (b) How much instantaneous power does she exert at t = 2.0 s ?

Answer 1

The work done by the girl on the wagon can be calculated using the formula Work = Force * Distance * cos(theta), where the force is 10 N and the angle is 37°. The instantaneous power exerted by the girl at t=2.0s can be found using the formula Power = Work / Time.

Step-by-step explanation:

(a) To find the work done by the girl on the wagon in the first 2.0 s, we can use the formula:

Work = Force * Distance * cos(theta)

where Force is the applied force, Distance is the displacement, and theta is the angle between the force and displacement vectors.

In this case, the applied force is 10 N, the distance is the displacement during the first 2 seconds, and the angle is 37°. So we can calculate:

Work = 10 N * Displacement * cos(37°)

(b) Instantaneous power is given by the formula:

Power = Work / Time

where Work is the work done and Time is the duration. So for t = 2.0 s, we can substitute the calculated work into the formula:

Power = Work / 2.0 s

Now we just need to perform the calculations to find the specific values.

Answer 2

The girl does 39.65 J of work on the wagon in the first 2.0 s. She exerts an instantaneous power of 47.25 W at t = 2.0 s.

Step-by-step explanation:

(a) How much work does the girl do on the wagon in the first 2.0 s?

To calculate the work done by the girl on the wagon, we need to find the displacement of the wagon and the angle between the applied force and the displacement. The work done is given by the equation:

Work = Force * Displacement * cos(θ)

In this case, the force is 10 N, and the displacement can be calculated using the formula for constant acceleration:

Displacement = (1/2) * acceleration * time^2

Since the wagon starts from rest, the initial velocity is 0 m/s. Using the equation of motion, v = u + at, we can find the acceleration:

0 = u + at, where u = 0, a = F / m, and t = 2 s

0 = 0 + (F / m) * 2

F = (m / 2) * a

Substituting the given values, we have:

F = (15 kg / 2) * a

10 N = (15 kg / 2) * a

a = (10 N * 2) / (15 kg)

a = 1.33 m/s^2

Now we can find the displacement:

Displacement = (1/2) * 1.33 m/s^2 * (2 s)^2

Displacement = 5.32 m

Substituting the values into the equation for work:

Work = 10 N * 5.32 m * cos(37°)

Work = 39.65 J

(b) How much instantaneous power does she exert at t = 2.0 s?

The instantaneous power is given by the equation:

Power = Force * velocity * cos(θ)

Using the equation of motion, v = u + at, we can find the final velocity:

v = u + at

v = 0 + 1.33 m/s^2 * 2 s

v = 2.66 m/s

Substituting the values into the equation for power:

Power = 10 N * 2.66 m/s * cos(37°)

Power = 47.25 W