A vehicle with a mass of 1200 kg accelerates uniformly from rest upwards with an incline of 1:25 and reaches a speed of 72 km/h after 2 minutes,calculate:

1.The kineyic energy of the vehicle after 2 minutes
2.The potential energy of the vehicle after 2 minutes

Hints:

Kinetic energy,
KE=(1/2)mv²
where
m=mass of object (kg)
v=velocity (m/s)

Potential energy
=mgh
m=mass in kg
g=acceleration due to gravity, m/s-2
h=height above datum in m.

To calculate h, you need the distance travelled.
For uniform acceleration, distance d, is the average speed (72 km/h ÷ 2, converted to m/s) multiplied by the time (2 minutes, converted to seconds).
The height, h is then
h=d*slope = d/25 (for a 1/25 slope)

To calculate the kinetic energy of the vehicle after 2 minutes, first we need to find its final velocity in m/s.

1 km/h = 0.2778 m/s
72 km/h = 72 * 0.2778 = 20 m/s

Now, we can calculate the kinetic energy using the formula:

Kinetic Energy = (1/2) * mass * velocity^2

1. The kinetic energy of the vehicle after 2 minutes is:

Kinetic Energy = (1/2) * 1200 kg * (20 m/s)^2
= 1/2 * 1200 * 400
= 240,000 J

Now, to calculate the potential energy of the vehicle after 2 minutes, we need to find its height above the ground first.

The incline of 1:25 means that for every 25 units in the horizontal direction, there is a 1 unit increase in height.

In this case, the ratio of horizontal distance traveled to vertical height gained is 25:1.

2. The potential energy of the vehicle after 2 minutes is given by:

Potential Energy = mass * gravitational acceleration * height

To find height, we need to find the horizontal distance covered in 2 minutes:

Speed = Distance / Time

72 km/h = Distance / 2 minutes

Distance = 72 km/h * (2/60) h/min * 1000 m/km
= 2400 m

Now, we can find the height gained:

Height = Distance / horizontal-to-vertical ratio
= 2400 m / 25
= 96 m

Now, we can calculate the potential energy using the formula:

Potential Energy = 1200 kg * 9.8 m/s^2 * 96 m
= 1,129,920 J

Therefore, the potential energy of the vehicle after 2 minutes is 1,129,920 Joules.

To find the kinetic energy of the vehicle after 2 minutes, we first need to calculate its final velocity.

Given:
Mass of the vehicle (m) = 1200 kg
Acceleration (a) = ?
Time (t) = 2 minutes = 2(60) = 120 seconds
Initial velocity (u) = 0 m/s (since it starts from rest)
Final velocity (v) = 72 km/h = 72 × (1000/3600) m/s = 20 m/s

We can find the acceleration using the formula:

v = u + at

Rearranging the formula to solve for acceleration:

a = (v - u) / t

Substituting the given values:

a = (20 - 0) / 120
a = 1/6 m/s²

Now, we can calculate the kinetic energy using the formula:

Kinetic energy (KE) = (1/2) × mass × velocity²

Substituting the given values:

KE = (1/2) × 1200 × (20)²
KE = 24000 J

Therefore, the kinetic energy of the vehicle after 2 minutes is 24,000 Joules.

To find the potential energy of the vehicle after 2 minutes, we will use the formula:

Potential energy (PE) = mass × gravity × height

Given:
Mass of the vehicle (m) = 1200 kg
Height of the incline (h) = ?

The incline is given as a ratio of 1:25, which means for every 1 unit of height, there are 25 units of horizontal distance. Let's assume the height of the incline is h.
By using the trigonometric relationship of an angle, we know that the height (opposite side) can be found using the ratio of the angle to the hypotenuse (distance), which is 1/25 in this case.

Using the tangent function:

tan(θ) = opposite/adjacent
1/25 = h/x (where x is the horizontal distance)

Rearranging the equation to solve for h:

h = (1/25) * x

To find x, we can use the distance formula:

distance (x) = velocity × time

Substituting the given values:

x = 20 m/s × 120 s
x = 2400 m

Now we can substitute x into the equation for h:

h = (1/25) * 2400
h = 96 m

Next, we can calculate the potential energy using the formula:

PE = mass × gravity × height

Assuming the acceleration due to gravity is 9.8 m/s²:

PE = 1200 kg × 9.8 m/s² × 96 m
PE = 1,132,160 J

Therefore, the potential energy of the vehicle after 2 minutes is 1,132,160 Joules.