5 kg cart is at the bottom of a 10 m hill. For the cart to reach the top of the hill, what is the minimum kinetic energy of the cart?
Assuming no friction:
KE = PE = mg*h = (5*9.8)*10=490 Joules.
To find the minimum kinetic energy required to bring the cart from the bottom to the top of the hill, we need to consider the conservation of energy.
The potential energy at the bottom of the hill will be converted into kinetic energy at the top of the hill.
Given:
Mass of the cart (m) = 5 kg
Height of the hill (h) = 10 m
The potential energy (PE) at the bottom of the hill is given by the formula:
PE = mgh
where:
m = mass (5 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height (10 m)
PE = mgh
= 5 kg * 9.8 m/s^2 * 10 m
= 490 joules
Since potential energy at the bottom of the hill will be converted into kinetic energy at the top, the minimum kinetic energy required for the cart to reach the top of the hill will be equal to the potential energy at the bottom of the hill.
Therefore, the minimum kinetic energy of the cart is 490 joules.
To find the minimum kinetic energy required for the cart to reach the top of the hill, we can consider the conservation of energy.
First, let's calculate the gravitational potential energy of the cart at the bottom of the hill. The gravitational potential energy is given by the formula:
Potential energy = mass * gravity * height
where,
mass = 5 kg (mass of the cart)
gravity = 9.8 m/s^2 (acceleration due to gravity)
height = 10 m (height of the hill)
Plugging in the values, we get:
Potential energy = 5 kg * 9.8 m/s^2 * 10 m
= 490 Joules
According to the conservation of energy, this potential energy at the bottom of the hill must be converted into kinetic energy at the top of the hill. Therefore, the minimum kinetic energy required for the cart to reach the top of the hill is 490 Joules.