A horizontal force of 50N is applied to a 2.0Kg trolly, initially at rest, and it moves a distance of 4.0m along a level, frictionless. The force then changes to 20N and acts for an aditional distance of 2.0m.
a) what is the the final kinetic energy of the trolley?
b)what is the trolley's final velocity?
total work done=50*4+20*2 joules
final KE= work done
a) Work(total)= Final KE = (50*4+20*2) J
= 240J
b)Ek = 1/2 mv(final)^2 - 1/2 mv(initial)^2
v(final)^2 = Ek / 1/2 x 2
= 240 J / 1
= 240
V(final)= 15.5 m/s
To find the final kinetic energy of the trolley, we can use the formula:
Kinetic energy (KE) = 1/2 * mass * velocity^2
Given:
Force (F1) = 50 N
Distance (d1) = 4.0 m
Force (F2) = 20 N
Distance (d2) = 2.0 m
Mass (m) = 2.0 kg
First, let's find the initial acceleration of the trolley using Newton's second law:
F1 = m * a1
50 N = 2.0 kg * a1
a1 = 50 N / 2.0 kg
a1 = 25 m/s^2
Next, let's find the velocity of the trolley after the first distance (d1):
v1^2 = u^2 + 2 * a1 * d1
Since the trolley is initially at rest (u = 0), the equation simplifies to:
v1^2 = 2 * a1 * d1
v1^2 = 2 * 25 m/s^2 * 4.0 m
v1^2 = 200 m^2/s^2
v1 ≈ 14.1 m/s (taking the square root of both sides)
Now, let's find the final acceleration of the trolley using Newton's second law again:
F2 = m * a2
20 N = 2.0 kg * a2
a2 = 20 N / 2.0 kg
a2 = 10 m/s^2
Finally, let's find the final velocity of the trolley using the initial velocity (v1) and the final acceleration (a2):
v2^2 = v1^2 + 2 * a2 * d2
v2^2 = (14.1 m/s)^2 + 2 * 10 m/s^2 * 2.0 m
v2^2 = 198.8 m^2/s^2 + 40 m^2/s^2
v2^2 ≈ 238.8 m^2/s^2
v2 ≈ 15.4 m/s (taking the square root of both sides)
a) The final kinetic energy of the trolley can be calculated using the final velocity (v2):
KE = 1/2 * m * v2^2
KE = 1/2 * 2.0 kg * (15.4 m/s)^2
KE ≈ 237.6 J
Therefore, the final kinetic energy of the trolley is approximately 237.6 Joules.
b) The trolley's final velocity is approximately 15.4 m/s.
To find the answers to these questions, we need to use the equations of motion and the principles of work and energy. Let's break down the problem step by step.
First, let's find the work done by the force of 50N during the first part of the motion. The work done is given by the equation:
Work = Force × Distance × Cos(θ)
Where:
- Force is the magnitude of the force applied (50N).
- Distance is the distance covered by the trolley in the first part of the motion (4.0m).
- θ is the angle between the applied force and the direction of motion (since the force is horizontal, θ is 0°, and Cos(0°) = 1).
Work = 50N × 4.0m × Cos(0°) = 200 Joules
The work done on an object is equal to the change in its kinetic energy. Therefore, the work done by this force increased the kinetic energy of the trolley by 200 Joules.
a) The final kinetic energy of the trolley is 200 Joules.
Now let's find the final velocity of the trolley.
We can use the work-energy principle, which states that the work done on an object equals the change in its kinetic energy:
Work = Change in Kinetic Energy
From the previous calculation, we know that the work done on the trolley is 200 Joules, which is also equal to the change in kinetic energy. Therefore:
200 Joules = (1/2) × Mass × (Final Velocity^2 - Initial Velocity^2)
The initial velocity of the trolley is zero, as stated in the problem. So the equation becomes:
200 Joules = (1/2) × 2.0Kg × (Final Velocity^2 - 0)
200 Joules = 1.0Kg × (Final Velocity^2)
Divide both sides of the equation by 1.0Kg:
200 Joules / 1.0Kg = Final Velocity^2
Final Velocity^2 = 200 Joules / 1.0Kg
Final Velocity^2 = 200 m^2/s^2
Taking the square root of both sides, we get:
Final Velocity = √(200 m^2/s^2) ≈ 14.14 m/s
b) The trolley's final velocity is approximately 14.14 m/s.
So, the final kinetic energy of the trolley is 200 Joules, and its final velocity is approximately 14.14 m/s.