An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. 𝐹1 is 25.1 N Acting due east, 𝐹2 37.0 N acting due north, 𝐹3 52.1 N acting due...

Question

An object, which is initially at rest on a frictionless horizontal surface, is acted upon by four constant forces. 𝐹1 is 25.1 N Acting due east, 𝐹2 37.0 N acting due north, 𝐹3 52.1 N acting due west, and 𝐹4 10.8 N acting due south. How much total work is done on the object in 4.44 s, if it has a mass of 16.0 kg? What is the final velocity after 4.44 s?

Which type of energy is changing for the object while the work is being done?

A) gravitational potential energy

B) kinetic energy

C) internal energy

D) elastic potential energy

Answer 1

To determine the total work done on the object, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy:

W = ΔK

Since the object starts at rest, its initial kinetic energy is zero, and we only need to calculate its final kinetic energy to determine the work done. To find the final velocity, we can use the equations of motion:

v = u + at

where u is the initial velocity (zero in this case), a is the acceleration, and t is the time.

To find the acceleration, we can use Newton's second law:

ΣF = ma

where ΣF is the vector sum of the forces acting on the object, m is the mass of the object, and a is its acceleration.

We can break each force into its x- and y-components:

F1x = 25.1 N (east) F2y = 37.0 N (north) F3x = -52.1 N (west) F4y = -10.8 N (south)

Note that the negative sign in front of F3x and F4y indicates that these forces are in the opposite direction to the positive x- and y-axes, respectively.

Summing the x- and y-components separately, we get:

ΣFx = F1x + F3x = 25.1 N - 52.1 N = -27.0 N (west)

ΣFy = F2y + F4y = 37.0 N - 10.8 N = 26.2 N (north)

The net force is then:

ΣF = √(ΣFx^2 + ΣFy^2) = √((-27.0 N)^2 + (26.2 N)^2) ≈ 37.0 N

The acceleration of the object is:

a = ΣF / m = 37.0 N / 16.0 kg ≈ 2.31 m/s^2

The final velocity of the object after 4.44 s is:

v = u + at = 0 + 2.31 m/s^2 × 4.44 s ≈ 10.3 m/s

The final kinetic energy of the object is:

K = (1/2)mv^2 = (1/2) × 16.0 kg × (10.3 m/s)^2 ≈ 853 J

Therefore, the total work done on the object is:

W = ΔK = K - 0 = 853 J

The type of energy that is changing for the object while the work is being done is kinetic energy (as seen in the work-energy principle above).